【LeetCode与《代码随想录》】二叉树篇：做题笔记与总结-JavaScript版

代码随想录

代码随想录
代码随想录CSDN官方

前、中、后指的都是根的位置。

144. 二叉树的前序遍历

https://leetcode.cn/problems/binary-tree-preorder-traversal/

var preorderTraversal = function (root) {
    // 前序遍历：中 左 右
    let ans = []

    const dfs = function (root) {
        if (root === null) return;
        ans.push(root.val);
        dfs(root.left);
        dfs(root.right);
    }

    dfs(root);

    return ans;
};

94. 二叉树的中序遍历

https://leetcode.cn/problems/binary-tree-inorder-traversal/

var inorderTraversal = function (root) {
    let ans = []
    const dfs = function (root) {
        if (root === null) return;
        // 中序:左中右
        dfs(root.left);
        ans.push(root.val);
        dfs(root.right);
    }
    dfs(root);
    return ans;
};

145. 二叉树的后序遍历

https://leetcode.cn/problems/binary-tree-postorder-traversal/

var postorderTraversal = function (root) {
    let ans = []
    const dfs = function (root) {
        if (root === null) return;
        // 后序:左右中
        dfs(root.left);
        dfs(root.right);
        ans.push(root.val);
    }
    dfs(root);
    return ans;
};

102.二叉树的层序遍历

https://leetcode.cn/problems/binary-tree-level-order-traversal/

var levelOrder = function (root) {
    let ans = [], queue = []
    if (root === null) return ans;
    queue.push(root);
    while (queue.length) {
        let len = queue.length;
        let anss = [];
        for (let i = 0; i < len; i++) {
            let node = queue.shift();
            anss.push(node.val);
            node.left && queue.push(node.left);
            node.right && queue.push(node.right);
        }
        ans.push(anss);
    }
    return ans;
};

226.翻转二叉树

https://leetcode.cn/problems/invert-binary-tree/

var invertTree = function (root) {
    const dfs = function (root) {
        if (root === null) return;
        dfs(root.left);
        dfs(root.right);
        [root.left, root.right] = [root.right, root.left];
    }
    dfs(root);
    return root;
};

101. 对称二叉树

https://leetcode.cn/problems/symmetric-tree/

var isSymmetric = function (root) {
    let ans = true;
    if (root === null) return ans;
    if (root.left && root.right === null) return false;
    if (root.right && root.left === null) return false;
    const dfs = function (l, r) {
        if (!l && !r) return;
        if (l && !r) {
            ans = false; return;
        }
        if (!l && r) {
            ans = false; return;
        }
        if (l.val !== r.val) {
            ans = false;
            return;
        }
        dfs(l.left, r.right);
        dfs(l.right, r.left);
    }

    dfs(root.left, root.right);
    return ans;
};

104.二叉树的最大深度

https://leetcode.cn/problems/maximum-depth-of-binary-tree/

var maxDepth = function (root) {
    let ans = 0;
    if (!root) return ans;
    let q = [];
    q.push(root);
    while (q.length) {
        ans++;
        let len = q.length;
        for (let i = 0; i < len; i++) {
            let node = q.shift();
            node.left && q.push(node.left);
            node.right && q.push(node.right);
        }
    }
    return ans;
};

111.二叉树的最小深度

https://leetcode.cn/problems/minimum-depth-of-binary-tree/

var minDepth = function (root) {
    // 若一个节点没有叶子节点，则结束
    let ans = 0;
    if (!root) return ans;
    let q = [];
    q.push(root);
    while (q.length) {
        ans++;
        let len = q.length;
        for (let i = 0; i < len; i++) {
            let node = q.shift();
            if (!node.left && !node.right) return ans;
            node.left && q.push(node.left);
            node.right && q.push(node.right);
        }
    }
};

222.完全二叉树的节点个数

https://leetcode.cn/problems/count-complete-tree-nodes/

注意：要利用完全二叉树的性质。

var countNodes = function (root) {
    if (!root) return 0;
    let l = root.left, r = root.right;
    let ll = 0, rr = 0;
    while (l) {
        ll++; l = l.left;
    }
    while (r) {
        rr++; r = r.right;
    }
    if (ll === rr) {
        return Math.pow(2, ll + 1) - 1;
    }
    else {
        return countNodes(root.left) + countNodes(root.right) + 1;
    }
};

110.平衡二叉树

https://leetcode.cn/problems/balanced-binary-tree/

var isBalanced = function (root) {
    var getHeight = function (root) {
        if (!root) return 0;
        let l = getHeight(root.left);
        if (l === -1) return -1;
        let r = getHeight(root.right);
        if (r === -1) return -1;

        if (Math.abs(l - r) > 1) return -1;
        else return Math.max(l, r) + 1;
    }

    let ans = getHeight(root);
    if (ans === -1) return false;
    else return true;
};

257. 二叉树的所有路径

https://leetcode.cn/problems/binary-tree-paths/

var binaryTreePaths = function (root) {
    let ans = [], anss = [];
    var dfs = function (root) {
        anss.push(root.val);
        root.left && dfs(root.left);
        root.right && dfs(root.right);
        if (!root.left && !root.right) {
            ans.push(anss.join('->'));
        }
        anss.pop();
    }

    dfs(root);
    return ans;
};

404.左叶子之和

https://leetcode.cn/problems/sum-of-left-leaves/

var sumOfLeftLeaves = function (root) {
    let ans = 0;
    var dfs = function (root, flag) {
        if (!root) return;
        root.left && dfs(root.left, 1);
        root.right && dfs(root.right, 0);
        if (flag === 1 && !root.left && !root.right) {
            ans += root.val;
        }
    }

    dfs(root);
    return ans;
};

513.找树左下角的值

https://leetcode.cn/problems/find-bottom-left-tree-value/

var findBottomLeftValue = function (root) {
    // 记录bfs每一层的第一个
    let ans = root.val, q = [];
    q.push(root);
    while (q.length) {
        let len = q.length;
        for (let i = 0; i < len; i++) {
            let node = q.shift();
            if (!i) ans = node.val;
            node.left && q.push(node.left);
            node.right && q.push(node.right);
        }
    }
    return ans;
};

112. 路径总和

https://leetcode.cn/problems/path-sum/

var hasPathSum = function (root, targetSum) {
    let ans = 0, flag = false;
    var dfs = function (root) {
        if (root.left) {
            ans += root.left.val;
            dfs(root.left);
            ans -= root.left.val;
        }
        if (root.right) {
            ans += root.right.val;
            dfs(root.right);
            ans -= root.right.val;
        }
        if (!root.left && !root.right) {
            if (ans === targetSum) flag = true;
        }
    }
    if (root) {
        ans += root.val;
        dfs(root);
    }


    return flag;
};

106.从中序与后序遍历序列构造二叉树

https://leetcode.cn/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

var buildTree = function (inorder, postorder) {
    if (!inorder.length) return null;
    let rootVal = postorder.pop();
    let rootIndex = inorder.indexOf(rootVal);
    let node = new TreeNode(rootVal);
    node.left = buildTree(inorder.slice(0, rootIndex), postorder.slice(0, rootIndex));
    node.right = buildTree(inorder.slice(rootIndex + 1), postorder.slice(rootIndex));
    return node;
};

105. 从前序与中序遍历序列构造二叉树

https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

var buildTree = function (preorder, inorder) {
    if (!preorder.length) return null;
    let rootVal = preorder.shift();
    let rootIndex = inorder.indexOf(rootVal);
    let node = new TreeNode(rootVal);
    node.left = buildTree(preorder.slice(0, rootIndex), inorder.slice(0, rootIndex));
    node.right = buildTree(preorder.slice(rootIndex), inorder.slice(rootIndex + 1));
    return node;
};

654.最大二叉树

https://leetcode.cn/problems/maximum-binary-tree/

var constructMaximumBinaryTree = function (nums) {
    // 数组的左右边界
    var dfs = function (l, r) {
        if (l > r) return null;
        let maxVal = -1, maxIndex = -1;
        for (let i = l; i <= r; i++) {
            if (nums[i] > maxVal) {
                maxVal = nums[i];
                maxIndex = i;
            }
        }
        const node = new TreeNode(maxVal);
        node.left = dfs(l, maxIndex - 1);
        node.right = dfs(maxIndex + 1, r);
        return node;
    }
    return dfs(0, nums.length - 1);
};

617.合并二叉树

https://leetcode.cn/problems/merge-two-binary-trees/

var mergeTrees = function (root1, root2) {
    // 参数:对应相同位置的两个节点
    var dfs = function (n1, n2) {
        if (!n1 && !n2) return null;
        if (n1 && !n2) return n1;
        if (n2 && !n1) return n2;
        let node = new TreeNode(n1.val + n2.val);
        node.left = dfs(n1.left, n2.left);
        node.right = dfs(n1.right, n2.right);
        return node;
    }
    return dfs(root1, root2);
};

700.二叉搜索树中的搜索

https://leetcode.cn/problems/search-in-a-binary-search-tree/

注意：要利用二叉搜索树的性质。

var searchBST = function (root, val) {
    let ans = null;
    var find = function (root) {
        if (root) {
            if (root.val === val) {
                ans = root;
                return;
            }
            if (val < root.val) find(root.left);
            if (val > root.val) find(root.right);
        }
    }
    find(root);
    return ans;
};

98.验证二叉搜索树

https://leetcode.cn/problems/validate-binary-search-tree/

注意：要判断每一个子树是否是二叉搜索树。

要用到二叉搜索树的性质：其中序遍历是一个递增的序列。

var isValidBST = function (root) {
    let arr = [];
    var dfs = function (root) {
        if (root) {
            dfs(root.left);
            arr.push(root.val);
            dfs(root.right);
        }
    }
    dfs(root);
    let ans = true;
    for (let i = 1; i < arr.length; i++) {
        if (arr[i] <= arr[i - 1]) {
            ans = false;
            break;
        }
    }
    return ans;
};

530.二叉搜索树的最小绝对差

https://leetcode.cn/problems/minimum-absolute-difference-in-bst/

var getMinimumDifference = function (root) {
    let arr = [];
    var dfs = function (root) {
        if (root) {
            dfs(root.left);
            arr.push(root.val);
            dfs(root.right);
        }
    }
    let ans = 100000 + 1;
    dfs(root);
    for (let i = 1; i < arr.length; i++) {
        let temp = Math.abs(arr[i] - arr[i - 1]);
        ans = Math.min(ans, temp);
    }
    return ans;
};

501.二叉搜索树中的众数

https://leetcode.cn/problems/find-mode-in-binary-search-tree/

用了额外的空间：

var findMode = function (root) {
    let arr = []
    var dfs = function (root) {
        if (root) {
            arr.push(root.val);
            root.left && dfs(root.left);
            root.right && dfs(root.right);
        }
    }
    dfs(root);
    let ans = new Map();
    for (let i = 0; i < arr.length; i++) {
        ans.set(arr[i], (ans.get(arr[i]) || 0) + 1);
    }
    let anss = Array.from(ans);
    anss.sort((a, b) => (b[1] - a[1]));
    // console.log(anss)
    let res = [], num = anss[0][1];
    for (let i = 0; i < anss.length; i++) {
        if (anss[i][1] === num) {
            res.push(anss[i][0]);
        }
        else {
            break;
        }
    }
    return res;
};

没有用额外的空间：

var findMode = function (root) {
    let ans = [], maxx = 0, nowVal = 100000 + 1, nowNum = 0;
    var dfs = function (root) {
        if (root) {
            root.left && dfs(root.left);
            if (nowVal === root.val) {
                nowNum++;

            }
            else {
                nowNum = 1;
                nowVal = root.val;
            }
            if (nowNum === maxx) {
                ans.push(root.val);
            }
            else if (nowNum > maxx) {
                ans = [];
                maxx = nowNum;
                ans.push(root.val);
            }
            root.right && dfs(root.right);
        }
    }
    dfs(root);
    return ans;
};

两项的差距还是比较大的，建议练习“不用额外空间”的版本。

236. 二叉树的最近公共祖先

https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-tree/

要找公共祖先，显然要从下往上找，即后序遍历。

var lowestCommonAncestor = function (root, p, q) {

    var dfs = function (root) {
        if (root === null || root === p || root === q) {
            return root;
        }
        let l = dfs(root.left);
        let r = dfs(root.right);

        if (l && r) {
            return root;
        } else if (r) {
            return r;
        } else if (l) {
            return l;
        } else {
            return null;
        }

    }
    return dfs(root);
};

235. 二叉搜索树的最近公共祖先

https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-search-tree/

• 要用二叉搜索树的性质：根节点与子树的大小关系
• 从根往下遍历，若根节点的值大于pq的值，则往左子树遍历，反之亦然
• 当第一次出现根的值在pq之间时，这就是最近的公共祖先
• 原因：
• 当出现根的值在pq之间时，说明pq一个在根的左子树，一个在根的右子树，若往左遍历会错过右子树的目标，若往右遍历会错过左子树的目标
• 有没有可能此时的根不是最近的祖先节点，而是次近的？答：绝无可能。在数值上是不符合搜索二叉树的性质的。次近的祖先节点不会在pq的值之间，只会比max(p,q)大，或比min(p,q)小

var lowestCommonAncestor = function (root, p, q) {
    let ans = null;
    var dfs = function (root) {
        if (root && !ans) {
            if (root.val >= p.val && root.val <= q.val) {
                ans = root; return;
            } else if (root.val >= q.val && root.val <= p.val) {
                ans = root; return;
            }

            if (root.val >= p.val && root.val >= q.val) {
                dfs(root.left);
            }
            if (root.val <= p.val && root.val <= q.val) {
                dfs(root.right);
            }
        }
    }
    dfs(root);
    return ans;
};

701.二叉搜索树中的插入操作

https://leetcode.cn/problems/insert-into-a-binary-search-tree/

var insertIntoBST = function (root, val) {
    let ans = false;
    if (root === null) {
        root = new TreeNode(val);
        return root;
    }
    var dfs = function (root) {
        if (root && !ans) {
            if (val > root.val) {
                // 往右
                if (root.right) dfs(root.right);
                else {
                    root.right = new TreeNode(val);
                    ans = true;
                    return;
                }
            }
            else if (val < root.val) {
                // 往左
                if (root.left) dfs(root.left);
                else {
                    root.left = new TreeNode(val)
                    ans = true;
                    return;
                }
            }
        }
    }

    dfs(root);
    return root;
};

450.删除二叉搜索树中的节点

https://leetcode.cn/problems/delete-node-in-a-bst/

注意：分类讨论。

• 要删除的key不存在
• 要删除的key：
• 是叶子节点（无左无右）
• 不是叶子节点但只有一个子树
• 不是叶子节点但有两个子树

注意根节点是key的情况。

var deleteNode = function (root, key) {
    // flag 1左节点删 2右节点删
    let find = false, parent = null, flag = 0;
    
    // 删除key节点
    var deletee = function (root, flag) {
        let node = null;
        if (flag === 1) {
            node = root.left;
        } else if (flag === 2) {
            node = root.right;
        }

        // 要删除节点为叶子节点
        if (!node.left && !node.right) {
            if (flag === 1) {
                root.left = null;
            } else if (flag === 2) {
                root.right = null;
            }
            return;
        }
        // 要删除的节点只有一个子节点
        else if (node.left && !node.right) {
            if (flag === 1) {
                root.left = node.left;
            } else if (flag === 2) {
                root.right = node.left;
            }
            return;
        }
        else if (node.right && !node.left) {
            if (flag === 1) {
                root.left = node.right;
            } else if (flag === 2) {
                root.right = node.right;
            }
            return;
        }
        // 要删除的节点有两个子节点
        else if (node.left && node.right) {
            // 右边接到左边
            let num = node.right.val, ans = false;

            var solve = function (roott) {
                if (roott && !ans) {
                    if (num > roott.val) {
                        if (roott.right) solve(roott.right);
                        else {
                            roott.right = node.right;
                            ans = true; return;
                        }
                    }
                    else if (num < roott.val) {
                        if (roott.left) solve(roott.left);
                        else {
                            roott.left = node.right;
                            ans = true; return;
                        }
                    }
                }
            }

            solve(node.left);
            if (flag === 1) {
                root.left = node.left;
            } else if (flag === 2) {
                root.right = node.left;
            }
        }
    }

    // 查找key节点是否存在
    var findKey = function (root) {
        if (root && !find) {
            if (root.left && root.left.val === key) {
                find = true;
                parent = root;
                flag = 1;
                return;
            }
            if (root.right && root.right.val === key) {
                find = true;
                parent = root;
                flag = 2;
                return;
            }
            findKey(root.left);
            findKey(root.right);
        }
    }

    // 根节点单独计算
    if (root && root.val === key) {
        // 左右子树都不为空
        if (root.left && root.right) {
            let ans = false, val = root.right.val;
            var dfs = function (node) {
                if (node && !ans) {
                    if (val > node.val) {
                        if (node.right) dfs(node.right);
                        else {
                            node.right = root.right;
                            ans = true; return;
                        }
                    }
                    if (val < node.val) {
                        if (node.left) dfs(node.left);
                        else {
                            node.left = root.right;
                            ans = true; return;
                        }
                    }
                }
            }
            dfs(root.left);
            return root.left;
        } else if (root.left) {
            return root.left;
        } else if (root.right) {
            return root.right;
        } else {
            return null;
        }
    }
    // 没找到
    findKey(root);
    if (find === false) return root;
    // 找到了
    deletee(parent, flag);
    return root;
};

上面代码写的乱七八糟的（像是半递归半模拟），下面是可读性更高的代码，来自《代码随想录》

其实就是用好递归。

var deleteNode = function (root, key) {
    if (!root) return null;
    if (root.val < key) {
        root.right = deleteNode(root.right, key);
        return root;
    }
    else if (root.val > key) {
        root.left = deleteNode(root.left, key);
        return root;
    } else {
        // 找到要删除的节点
        if (!root.left && !root.right) return null;
        else if (root.left && !root.right) return root.left;
        else if (root.right && !root.left) return root.right;
        else {
            // 左右节点都在
            let rightNode = root.right;
            let minNodee = minNode(rightNode);
            root.val = minNodee.val;
            root.right = deleteNode(root.right, minNodee.val);
            return root;
        }
    }
};

var minNode = function (root) {
    while (root.left) {
        root = root.left;
    }
    return root;
}

669. 修剪二叉搜索树

https://leetcode.cn/problems/trim-a-binary-search-tree/

题解

var trimBST = function (root, low, high) {
    if (root === null) return null;
    if (root.val > high) return trimBST(root.left, low, high);
    if (root.val < low) return trimBST(root.right, low, high);
    root.left = trimBST(root.left, low, high);
    root.right = trimBST(root.right, low, high);
    return root;
};

108.将有序数组转换为二叉搜索树

https://leetcode.cn/problems/convert-sorted-array-to-binary-search-tree/

var sortedArrayToBST = function (nums) {
    var dfs = function (l, r) {
        if (l > r) return null;
        let mid = Math.floor(l + (r - l) / 2);
        let root = new TreeNode(nums[mid]);
        root.left = dfs(l, mid - 1);
        root.right = dfs(mid + 1, r);
        return root;
    }
    return dfs(0, nums.length - 1);
};

538.把二叉搜索树转换为累加树

https://leetcode.cn/problems/convert-bst-to-greater-tree/

把二叉搜索树转为数组，数组累加后再赋值给二叉树。

var convertBST = function (root) {
    if (root === null) return null;
    let arr = [];
    var getNum = function (root) {
        if (root) {
            getNum(root.left);
            arr.push(root.val);
            getNum(root.right);
        }
    }

    getNum(root);
    for (let i = arr.length - 2; i >= 0; i--) {
        arr[i] += arr[i + 1];
    }

    var solve = function (root) {
        if (root) {
            solve(root.left);
            root.val = arr[i++];
            solve(root.right);
            return root;
        }
    }
    let i = 0;
    return solve(root);
};

更简洁的方法：累加的顺序是右中左，按照此顺序递归累加即可。

var convertBST = function (root) {
    // 右中左
    if (!root) return null;
    let ans = 0;
    var dfs = function (root) {
        if (root) {
            dfs(root.right);
            ans += root.val;
            root.val = ans;
            dfs(root.left);
            return root;
        }
    }
    return dfs(root);
};

相关题目

107.二叉树的层次遍历II

https://leetcode.cn/problems/binary-tree-level-order-traversal-ii/

var levelOrderBottom = function (root) {
    let ans = [], queue = []
    if (root === null) return [];
    queue.push(root);
    while (queue.length) {
        let len = queue.length;
        let anss = []
        for (let i = 0; i < len; i++) {
            let node = queue.shift();
            anss.push(node.val);
            node.left && queue.push(node.left);
            node.right && queue.push(node.right);
        }
        ans.unshift(anss);
    }
    return ans;
};

199.二叉树的右视图

https://leetcode.cn/problems/binary-tree-right-side-view/

var rightSideView = function (root) {
    // 每一层的最后一个
    let ans = [], queue = [];
    if (root === null) return [];
    queue.push(root);
    while (queue.length) {
        let len = queue.length;
        for (let i = 0; i < len; i++) {
            let node = queue.shift();
            node.left && queue.push(node.left);
            node.right && queue.push(node.right);
            if (i === len - 1) {
                ans.push(node.val);
            }
        }
    }
    return ans;
};