golang 解法, 执行用时: 9 ms, 内存消耗: 6.1 MB, 提交时间: 2024-02-20 10:15:53

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findTilt(root *TreeNode) int {
    ans := 0
    var dfs func(*TreeNode) int
    dfs = func(node *TreeNode) int {
        if(node == nil){
            return 0
        }
        l := dfs(node.Left)
        r := dfs(node.Right)
        if v := l - r ; v > 0 {
            ans += v
        } else {
            ans -= v
        }
        return l + r + node.Val
    }
    dfs(root)
    return ans
}

javascript 解法, 执行用时: 67 ms, 内存消耗: 54 MB, 提交时间: 2024-02-20 10:15:32

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var findTilt = function(root) {
    let ans = 0;
    function dfs(node) {
        if(node == null)
            return 0;
        const l = dfs(node.left), r = dfs(node.right);
        ans += Math.abs(l - r);
        return node.val + l + r;
    }
    dfs(root);
    return ans;
};

java 解法, 执行用时: 0 ms, 内存消耗: 43.2 MB, 提交时间: 2024-02-20 10:15:08

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int ans;
    public int findTilt(TreeNode root) {
        ans = 0;
        dfs(root);
        return ans;
    }

    private int dfs(TreeNode node) {
        if(node == null)
            return 0;
        int l = dfs(node.left), r = dfs(node.right);
        ans += Math.abs(r - l);
        return l + r + node.val;
    }
}

python3 解法, 执行用时: 40 ms, 内存消耗: 17.8 MB, 提交时间: 2024-02-20 10:14:50

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def findTilt(self, root: TreeNode) -> int:
        def dfs(node):
            if not node:
                return 0, 0
            l_sum, l_diff = dfs(node.left)
            r_sum, r_diff = dfs(node.right)
            return l_sum + r_sum + node.val, l_diff + r_diff + abs(r_sum - l_sum)
        return dfs(root)[1]

golang 解法, 执行用时: 32 ms, 内存消耗: 6.4 MB, 提交时间: 2021-06-07 11:47:43

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findTilt(root *TreeNode) int {
    if root == nil {
        return 0
    }
    
    return getMax(sum(root.Left), sum(root.Right)) + findTilt(root.Left) + findTilt(root.Right)
}

// 所有节点之和
func sum(root *TreeNode) int {
    if root == nil {
        return 0
    }
    return root.Val + sum(root.Left) + sum(root.Right)
}

func getMax(a, b int) int {
    if a > b {
        return a - b
    }
    return b - a 
}

golang 解法, 执行用时: 28 ms, 内存消耗: 6.4 MB, 提交时间: 2021-06-07 11:31:47

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findTilt(root *TreeNode) int {
    if root == nil {
        return 0
    }
    
    return getMax(sum(root.Left), sum(root.Right)) + findTilt(root.Left) + findTilt(root.Right)
}

func sum(root *TreeNode) int {
    if root == nil {
        return 0
    }
    return root.Val + sum(root.Left) + sum(root.Right)
}

func getMax(a, b int) int {
    if a > b {
        return a - b
    }
    return b - a 
}