题目

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

For example:
Given the below binary tree,

       1
/ \
2 3

Return 6.

题解

递归求解。

取当前点和左右边加和,当前点的值中最大的作为本层返回值。并在全局维护一个max。使用数组,因为是引用类型。所以在递归过程中可以保存结果。

代码如下:

 1     public int maxPathSum(TreeNode root) {
 2         int[] max = new int[1];
 3         max[0] = Integer.MIN_VALUE;
 4         findmax(root,max);
 5         return max[0];
 6     }
 7     
 8     public int findmax(TreeNode root, int[] max){
 9         if(root==null)
             return 0;
         
         int left = findmax(root.left,max);
         int right = findmax(root.right,max);
         
         int ans = Math.max(root.val,Math.max(root.val+left, root.val+right));
         
         max[0] = Math.max(max[0],Math.max(ans,root.val+left+right));
         
         return ans;
         
     }

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