代码随想录刷题第17天

第一题是平衡二叉树https://leetcode.cn/problems/balanced-binary-tree/，初步思路是利用递归算node->left与node->right的差。卡在了如何表示node->left与node->right之差。递归返回值记录左右子树的平衡情况，平衡return height，失衡return -1，终止条件为node == NULL，return 0，单层递归的逻辑为判断左右子树高度之差，>1 return -1，否则return高度。

class Solution {
public:
    int getheight(TreeNode* node) {
        if (node == NULL)
            return 0;
        int leftheight = getheight(node->left);
        if (leftheight == -1)
            return -1;
        int rightheight = getheight(node->right);
        if (rightheight == -1)
            return -1;
        if (abs(leftheight - rightheight) > 1)
            return -1;
        else
            return max(leftheight, rightheight) + 1;
    }
    bool isBalanced(TreeNode* root) {
        int result = getheight(root);
        if (result == -1)
            return false;
        else
            return true;
    }
};

另外一种方法是迭代法。

class Solution {
private:
    int getDepth(TreeNode* cur) {
        stack st;
        if (cur != NULL)
            st.push(cur);
        int depth = 0;
        int result = 0;
        while (!st.empty()) {
            TreeNode* node = st.top();
            if (node != NULL) {
                st.pop();
                st.push(node);
                st.push(NULL);
                depth++;
                if (node->right)
                    st.push(node->right);
                if (node->left)
                    st.push(node->left);
            } else {
                st.pop();
                node = st.top();
                st.pop();
                depth--;
            }
            result = result > depth ? result : depth;
        }
        return result;
    }

public:
    bool isBalanced(TreeNode* root) {
        stack st;
        if (root == NULL)
            return true;
        st.push(root);
        while (!st.empty()) {
            TreeNode* node = st.top();
            st.pop();
            if (abs(getDepth(node->left) - getDepth(node->right)) > 1)
                return false;
            if (node->right)
                st.push(node->right);
            if (node->left)
                st.push(node->left);
        }
        return true;
    }
};

第二题是二叉树的所有路径https://leetcode.cn/problems/binary-tree-paths/，由于此题需要寻找父节点到叶子结点的路径，应此应先处理中间结点，用前序遍历的方式是最合适的。

class Solution {
public:
    void travelsal(TreeNode* cur, vector& path, vector& result) {
        path.push_back(cur->val);
        if (cur->left == NULL && cur->right == NULL) {
            string sPath;
            for (int i = 0; i < path.size() - 1; i++) {
                sPath += to_string(path[i]);
                sPath += "->";
            }
            sPath += to_string(path[path.size() - 1]);
            result.push_back(sPath);
            return;
        }
        if (cur->left) {
            travelsal(cur->left, path, result);
            path.pop_back();
        }
        if (cur->right) {
            travelsal(cur->right, path, result);
            path.pop_back();
        }
    }
    vector binaryTreePaths(TreeNode* root) {
        vector result;
        vector path;
        if (root == NULL)
            return result;
        travelsal(root, path, result);
        return result;
    }
};

下面是迭代法的代码。

class Solution {
public:
    vector binaryTreePaths(TreeNode* root) {
        stack treeSt;
        stack pathSt;
        vector result;
        if (root == NULL)
            return result;
        treeSt.push(root);
        pathSt.push(to_string(root->val));
        while (!treeSt.empty()) {
            TreeNode* node = treeSt.top();
            treeSt.pop();
            string path = pathSt.top();
            pathSt.pop();
            if (node->left == NULL && node->right == NULL) {
                result.push_back(path);
            }
            if (node->right) {
                treeSt.push(node->right);
                pathSt.push(path + "->" + to_string(node->right->val));
            }
            if (node->left) {
                treeSt.push(node->left);
                pathSt.push(path + "->" + to_string(node->left->val));
            }
        }
        return result;
    }
};

第三题是左叶子之和https://leetcode.cn/problems/sum-of-left-leaves/description/，递归代码如下：

class Solution {
public:
    int travel(TreeNode* node) {//后序遍历
        if (node == NULL)
            return 0;
        if (node->left == NULL && node->right == NULL)
            return 0;
        int leftnum = travel(node->left);
        if (node->left != NULL && node->left->left == NULL &&
            node->left->right == NULL) {
            leftnum = node->left->val;
        }
        int rightnum = travel(node->right);
        return leftnum + rightnum;
    }
    int sumOfLeftLeaves(TreeNode* root) { return travel(root); }
};

已经第17天了，加油加油！