124. 二叉树中的最大路径和(Binary Tree Maximum Path Sum)H

英文题目

A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.
The path sum of a path is the sum of the node's values in the path.
Given the root of a binary tree, return the maximum path sum of any path.
Example 1:

Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.

Example 2:

Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.

Constraints:
The number of nodes in the tree is in the range [1, 3 * 104].
1000 <= Node.val <= 1000

中文题目

路径被定义为一条从树中任意节点出发，沿父节点-子节点连接，达到任意节点的序列。同一个节点在一条路径序列中至多出现一次。该路径至少包含一个节点，且不一定经过根节点。
路径和是路径中各节点值的总和。
给你一个二叉树的根节点 root ，返回其最大路径和。
示例 1：

输入：root = [1,2,3]
输出：6
解释：最优路径是 2 -> 1 -> 3 ，路径和为 2 + 1 + 3 = 6

示例 2：

输入：root = [-10,9,20,null,null,15,7]
输出：42
解释：最优路径是 15 -> 20 -> 7 ，路径和为 15 + 20 + 7 = 42

提示：
树中节点数目范围是 [1, 3 * 10^4]
-1000 <= Node.val <= 1000

递归

递归内其实是后序遍历，其返回值是根节点加上最大值的左/右子树节点和
题目要求的是最大路径和，有三种情况
```
    a
   / \
  b   c
```
1.b+a+c，是代码中的lmr
2.b+a+a的父节点
3.c+a+a的父节点
2和3可合并起来，即 max(b,c) + a + a的父节点，
需注意的是，节点有可能是负值，最大和要想办法舍弃负值
然后每一次都把最大值保存起来
但返回的是根节点加上最大值的左/右子树的节点和

时间复杂度O(N)，空间复杂度(N)

# python3: 时间 76 ms, 击败 88.52%; 内存 24.4 MB, 击败 32.24%
# 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 maxPathSum(self, root: Optional[TreeNode]) -> int:
        self.res = float('-inf')
        self.dfs(root)
        return self.res
    
    def dfs(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        left = max(self.dfs(root.left), 0)
        right = max(self.dfs(root.right), 0)
        lmr = root.val + left + right
        self.res = max(self.res, lmr)
        return root.val + max(left, right)

// c++: 时间 12 ms, 击败 99.49%; 内存 27 MB, 击败 41.94%
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode* root) {
        int res = numeric_limits<int>::lowest();
        dfs(res, root);
        return res;
    }

    int dfs(int & res, TreeNode* root) {
        if (root == nullptr) {
            return 0;
        }
        int left = max(dfs(res, root->left), 0);
        int right = max(dfs(res, root->right), 0);
        int lmr = root->val + left + right;
        res = max(res, lmr);
        return root->val + max(left, right);;
    }
};

// java: 时间 0 ms, 击败 100%; 内存 42.6 MB, 击败 73.64%
/**
 * 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 {
    int res = Integer.MIN_VALUE;

    public int maxPathSum(TreeNode root) {
        dfs(root);
        return res;
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = Math.max(dfs(root.left), 0);
        int right = Math.max(dfs(root.right), 0);
        int lmr = root.val + left + right;
        res = Math.max(res, lmr);
        return root.val + Math.max(left, right);
    }
}

// go: 时间 12 ms, 击败 91.92%; 内存 6.9 MB, 击败 66.54%
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func maxPathSum(root *TreeNode) int {
    res := math.MinInt
    dfs(&res, root)
    return res
}

func dfs(res *int, root *TreeNode) int {
    if root == nil {
        return 0
    }
    left := MaxInt(dfs(res, root.Left), 0)
    right := MaxInt(dfs(res, root.Right), 0)
    lmr := root.Val + left + right
    *res = MaxInt(*res, lmr)
    return root.Val + MaxInt(left, right)
}

func MaxInt(x, y int) int {
    if x > y {
        return x
    }
    return y
}

// javascript: 时间 80 ms, 击败 54.78%; 内存 50.3 MB, 击败 74.77%
/**
 * 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 maxPathSum = function(root) {
    let res = Number.MIN_SAFE_INTEGER;
    
    const dfs = function(root) {
        if (root == null) {
            return 0;
        }
        let left = Math.max(dfs(root.left), 0);
        let right = Math.max(dfs(root.right), 0);
        let lmr = root.val + left + right;
        res = Math.max(res, lmr);
        return root.val + Math.max(left, right);
    };

    dfs(root);
    return res;
};

# 124. 二叉树中的最大路径和(Binary Tree Maximum Path Sum)H

# 英文题目

# 中文题目

# 递归

124. 二叉树中的最大路径和(Binary Tree Maximum Path Sum)H

英文题目

中文题目

递归