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算法笔记: 力扣#563 二叉树的坡度

admin 11月 12, 2020 0

问题描述

解法

分析

Python 实现

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class :
    def findTilt(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        self._tilt = 0
        self.dfs(root)
        return self._tilt
        
    def dfs(self, node):
        if node is None:
            return 0
        left = self.dfs(node.left)
        right = self.dfs(node.right)
        self._tilt += abs(left-right)
        return left+right+node.val

Java 实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class  {
    private int tilt;
    private int dfs(TreeNode node){
        if(node == null){ return 0; }
        int left = dfs(node.left);
        int right = dfs(node.right);
        tilt += Math.abs(left-right);
        return left+right+node.val;
    }
    public int findTilt(TreeNode root) {
        tilt = 0;
        dfs(root);
        return tilt;
    }
}

时间复杂度

O(n).

空间复杂度

O(n).

链接

563. Binary Tree Tilt
563. 二叉树的坡度
 (English version) Algorithm Notes: Leetcode#563 Binary Tree Tilt