[LeetCode] Binary Tree Tilt

Given a binary tree, return the tilt of the whole tree.

The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.

The tilt of the whole tree is defined as the sum of all nodes‘ tilt.

Example:

Input:
         1
       /         2     3
Output: 1
Explanation:
Tilt of node 2 : 0
Tilt of node 3 : 0
Tilt of node 1 : |2-3| = 1
Tilt of binary tree : 0 + 0 + 1 = 1

Note:

  1. The sum of node values in any subtree won‘t exceed the range of 32-bit integer.
  2. All the tilt values won‘t exceed the range of 32-bit integer.

使用后序遍历递归计算每一个节点的倾斜度,维护一个节点与子树和作为返回值返回上层递归调用。最后可得整棵树的倾斜度。

class Solution {
public:
    int res = 0;
    int findTilt(TreeNode* root) {
        dfs(root);
        return res;
    }
    int dfs(TreeNode* root) {
        if (root == nullptr)
            return 0;
        int left = dfs(root->left);
        int right = dfs(root->right);
        res += abs(left - right);
        return left + right + root->val;
    }
};
// 22 ms
时间: 2024-12-09 00:53:41

[LeetCode] Binary Tree Tilt的相关文章

[LeetCode] Binary Tree Tilt 二叉树的坡度

Given a binary tree, return the tilt of the whole tree. The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0. The tilt of the 

[leetcode]Binary Tree Zigzag Level Order Traversal

Binary Tree Zigzag Level Order Traversal Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example:Given binary tree {3,9,20,#

[leetcode]Binary Tree Level Order Traversal @ Python

原题地址:http://oj.leetcode.com/problems/binary-tree-level-order-traversal/ 题意:二叉树的层序遍历的实现. 解题思路:二叉树的层序遍历可以用bfs或者dfs来实现.这里使用的dfs实现,代码比较简洁.实际上,二叉树的先序遍历就是dfs实现.   比如一棵树如下: 1 /  \ 2       3 /    \    /   \ 4     5  6    7    二叉树的先序遍历为{1,2,4,5,3,6,7},可以看到这个遍

[LeetCode] Binary Tree Zigzag Level Order Traversal(bfs)

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / 9 20 / 15 7 return its zig

[leetcode]Binary Tree Zigzag Level Order Traversal @ Python

原题地址:http://oj.leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ 题意: Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between

[leetcode]Binary Tree Level Order Traversal II @ Python

原题地址:http://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/ 题意: Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary

[LeetCode] Binary Tree Zigzag Level Order Traversal 二叉树的之字形层序遍历

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / 9 20 / 15 7 return its zigz

LeetCode "Binary Tree *-order Traversal' by Iteration

Binary Tree *-order traversal by recursion is trivial. But their iteration version deserves a look: Pre-Order class Solution { vector<int> ret; public: vector<int> preorderTraversal(TreeNode *p) { if (!p) return ret; stack<TreeNode *> st

LeetCode: Binary Tree Maximum Path Sum [124]

[题目] Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example: Given the below binary tree, 1 / 2 3 Return 6. [题意] 给定一棵二叉树,找出其中路径和最大的路径,然会返回最大路径和. 本题中的路径不是从根节点到叶子节点这样的传统的路径,而是指的二叉树中任意两个节点之间的联通路径.