[LeetCode] 124. Binary Tree Maximum Path Sum 求二叉树的最大路径和

Given a non-empty binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

Example 1:

Input: [1,2,3]

       1
      / 
     2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / 
  9  20
    /  
   15   7

Output: 42

给一个非空二叉树，找出最大路径和。

解法：递归，与常规path sum不同，这题的path sum可以不起始于root，也可以不终止于leaf。

Java:

public class Solution {
    int maxValue;
    
    public int maxPathSum(TreeNode root) {
        maxValue = Integer.MIN_VALUE;
        maxPathDown(root);
        return maxValue;
    }
    
    private int maxPathDown(TreeNode node) {
        if (node == null) return 0;
        int left = Math.max(0, maxPathDown(node.left));
        int right = Math.max(0, maxPathDown(node.right));
        maxValue = Math.max(maxValue, left + right + node.val);
        return Math.max(left, right) + node.val;
    }
}

Python:

# Time:  O(n)
# Space: O(h), h is height of binary tree

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution(object):
    maxSum = float("-inf")

    # @param root, a tree node
    # @return an integer
    def maxPathSum(self, root):
        self.maxPathSumRecu(root)
        return self.maxSum

    def maxPathSumRecu(self, root):
        if root is None:
            return 0
        left = max(0, self.maxPathSumRecu(root.left))
        right = max(0, self.maxPathSumRecu(root.right))
        self.maxSum = max(self.maxSum, root.val + left + right)
        return root.val + max(left, right)　

C++:

class Solution {
public:
    int maxPathSum(TreeNode* root) {
        int res = INT_MIN;
        helper(root, res);
        return res;
    }
    int helper(TreeNode* node, int& res) {
        if (!node) return 0;
        int left = max(helper(node->left, res), 0);
        int right = max(helper(node->right, res), 0);
        res = max(res, left + right + node->val);
        return max(left, right) + node->val;
    }
};

　　

Follow up: 返回最大路径，递归函数就不能返回路径和，而是返回该路径上所有结点组成的数组，递归的参数还要保留最大路径和，对左右子节点调用递归函数后得到的是数组，统计出数组之和，并且跟0比较，如果小于0，和清零，数组清空。

类似题目：

[LeetCode] 112. Path Sum 路径和

[LeetCode] 113. Path Sum II 路径和 II

[LeetCode] 437. Path Sum III 路径和 III 

[LeetCode] 666. Path Sum IV 二叉树的路径和 IV

[LeetCode] 687. Longest Univalue Path 最长唯一值路径

[LeetCode] 129. Sum Root to Leaf Numbers 求根到叶节点数字之和

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