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  • 124 Binary Tree Maximum Path Sum

    给出一棵二叉树,寻找一条路径使其路径和最大。
    对于这个问题,路径被定义为从树中任意节点连接任意节点的序列。该路径必须至少包含一个节点,并且不需要经过根节点。
    例如:
    给出一棵二叉树:
           1
          /
         2   3
    返回 6。
    详见:https://leetcode.com/problems/binary-tree-maximum-path-sum/description/

    Java实现:

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        int maxSum=Integer.MIN_VALUE;
        public int maxPathSum(TreeNode root) {
            helper(root);
            return maxSum;
        }
        
        public int helper(TreeNode root){
            if(root==null){
                return 0;
            }
            int left = helper(root.left);
            int right = helper(root.right);
            int res = Math.max(root.val,Math.max(root.val+left, root.val+right));
            
            maxSum = Math.max(maxSum,Math.max(res,root.val+left+right));
            
            return res;
        }
    }
    

    参考:https://www.cnblogs.com/springfor/p/3886411.html

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  • 原文地址:https://www.cnblogs.com/xidian2014/p/8723034.html
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