代码随想录算法训练营第十八天|513 找树左下角的值 112 路径总和 113 路径总和|| 106 从中序和后序遍历序列构造二叉树

目录

513 找树左下角的值

迭代

递归

112 路径总和

迭代

递归

113 路径总和 II

106 从中序与后序遍历序列构造二叉树

105 从前序与中序遍历序列构造二叉树

---

513 找树左下角的值

迭代

class Solution {
    public int findBottomLeftValue(TreeNode root) {
        int res = 0;
        Deque st = new LinkedList<>();
        st.add(root);
        while(!st.isEmpty()){
            int siz = st.size();
            for(int i = 0;i < siz;i++){
                TreeNode cur = st.pop();
                if(i == 0)res = cur.val;
                if(cur.left != null)st.add(cur.left);
                if(cur.right != null)st.add(cur.right);
            }
        }
        return res;
    }
}

时间复杂度O(n)

空间复杂度O(n)

class Solution {
    public int findBottomLeftValue(TreeNode root) {
        int res = 0;
        Dequest = new LinkedList<>();
        st.add(root);
        while(!st.isEmpty()){
            TreeNode cur = st.pop();
            if(cur.right != null)st.add(cur.right);
            if(cur.left != null)st.add(cur.left);//确保左边最后被遍历到
            res = cur.val; 
        }
        return res;
    }
}

时间复杂度O(n)

空间复杂度O(n)

递归

class Solution {
    private int value = 0;
    private int depth = 0;
    public int findBottomLeftValue(TreeNode root) {
        value = root.val;
        findLeftValue(root,0);
        return value;
    }
    private void findLeftValue(TreeNode root,int dep){
        if(root.left == null && root.right == null){
            if(dep > depth){
                depth = dep;
                value = root.val;
            }
        }
        if(root.left != null)findLeftValue(root.left,dep + 1);
        if(root.right != null)findLeftValue(root.right,dep + 1);
    }
}

时间复杂度O(n)

空间复杂度O(n)

112 路径总和

迭代

class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if(root == null)return false;
        Dequest1 = new LinkedList<>();
        Dequest2 = new LinkedList<>();
        st1.add(root);
        st2.add(root.val);
        while(!st1.isEmpty()){
            int siz = st1.size();
            for(int i = 0;i < siz;i++){
                int sum = st2.pop();
                TreeNode cur = st1.pop();
                if(cur.left == null && cur.right == null && sum == targetSum)return true;
                if(cur.right != null){
                    st1.add(cur.right);
                    st2.add(cur.right.val + sum);
                }
                if(cur.left != null){
                    st1.add(cur.left);
                    st2.add(cur.left.val + sum);
                }
            }
        }
        return false;
    }
}

时间复杂度O(n)

空间复杂度O(n)

class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if(root == null)return false;
        Dequest1 = new LinkedList<>();
        Dequest2 = new LinkedList<>();
        st1.add(root);
        st2.add(root.val);
        while(!st1.isEmpty()){
            TreeNode cur = st1.pop();
            int sum = st2.pop();
            if(cur.left == null && cur.right == null && sum == targetSum)return true;
            if(cur.right != null){
                st1.add(cur.right);
                st2.add(cur.right.val + sum);
            }
            if(cur.left != null){
                st1.add(cur.left);
                st2.add(cur.left.val + sum);
            }
        }
        return false;
    }
}

时间复杂度O(n)

空间复杂度O(n)

递归

class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if(root == null)return false;
        if(root.left == null && root.right == null)return root.val == targetSum;
        return hasPathSum(root.left,targetSum - root.val) || hasPathSum(root.right,targetSum - root.val);
    }
}

时间复杂度O(n)

空间复杂度O(n)

113 路径总和 II

class Solution {
        List>res = new ArrayList<>();
        Listpath = new LinkedList<>();
    public List> pathSum(TreeNode root, int targetSum) {
        if(root == null)return res;
        dfs(root,targetSum);
        return res;
    }
    private void dfs(TreeNode root,int targetSum){
        path.add(root.val);
        if(root.left == null && root.right == null){
            if(root.val == targetSum){
                res.add(new ArrayList<>(path));
            }
            return;
        }
        if(root.left != null){
            dfs(root.left,targetSum - root.val);
            path.remove(path.size() - 1);
        }
        if(root.right != null){
            dfs(root.right,targetSum - root.val);
            path.remove(path.size() - 1);
        }
    }
}

时间复杂度O(n^2)

空间复杂度O(n)

106 从中序与后序遍历序列构造二叉树

class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder.length == 0 || inorder == null)return null;
        return buildHelper(inorder,0,inorder.length,postorder,0,postorder.length);
    }
    private TreeNode buildHelper(int[] inorder,int inorderFirst,int inorderEnd,int[] postorder,int postorderFirst,int postorderEnd){
        if(postorderFirst == postorderEnd)return null;
        int val = postorder[postorderEnd - 1];
        TreeNode cur = new TreeNode(val);
        int mid = inorderFirst;
        for(;mid < inorderEnd;mid++){
            if(inorder[mid] == cur.val)break;
        }
        //将inorder分为左部分，右部分
        int leftinorderFirst = inorderFirst;
        int leftinorderEnd = mid;
        int rightinorderFirst = mid + 1;
        int rightinorderEnd = inorderEnd;
        //将postorder分为左部分，右部分
        int leftpostorderFirst = postorderFirst;
        int leftpostorderEnd =  leftpostorderFirst + mid - inorderFirst;
        int rightpostorderFirst = leftpostorderEnd;
        int rightpostorderEnd = postorderEnd - 1;
        cur.left = buildHelper(inorder,leftinorderFirst,leftinorderEnd,postorder,leftpostorderFirst,leftpostorderEnd);
        cur.right = buildHelper(inorder,rightinorderFirst,rightinorderEnd,postorder,rightpostorderFirst,rightpostorderEnd);
        return cur;
    }
}

时间复杂度O(n)

空间复杂度O(n)

105 从前序与中序遍历序列构造二叉树