Binary Tree Maximum Path Sum

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.

注意： The path may start and end at any node in the tree. 而不是从叶子到叶子， 所以有很多种情况。

根本的考虑是递归， tree的结构首先考虑递归。

这里从最下面一层开始一层层往上走，直到root。查看每一层的max的值。

每一个root有两边，left and right， 分别查这两边的最值，然后比较，返回的是经过这个node的路径的最大值。

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     int max;
12     public int maxPathSum(TreeNode root) {
13         // Start typing your Java solution below
14         // DO NOT write main() function
15         max = root.val;
16         if(root.right == null && root.left == null) return root.val;
17         PathForNode(root);
18         return max;
19     }
20     public int PathForNode(TreeNode n){
21         if(n == null) return 0;
22         else{
23             int right = PathForNode(n.right);
24             int left = PathForNode(n.left);
25             if(max < n.val + right + left) max = n.val + right + left;
26             else if(max < n.val + left) max = n.val + left;
27             else if(max < n.val + right) max = n.val + right;
28             else if(max < n.val) max = n.val;
29             return max(n.val + right , n.val + left,n.val);
30             }
31     }
32     public int max(int a,int b,int c){
33         if(a < 0 && b < 0 && c < 0) return 0;
34         int s = (a < b) ? b : a;
35         return (s < c) ? c : s;
36     }
37 }

 第二遍：

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     int result = Integer.MIN_VALUE;
12     public int maxPathSum(TreeNode root) {
13         // IMPORTANT: Please reset any member data you declared, as
14         // the same Solution instance will be reused for each test case.
15         result = Integer.MIN_VALUE;
16         dfs(root);
17         return result;
18     }
19     public int dfs(TreeNode root){
20         if(root == null) return 0;
21         int left = dfs(root.left);
22         int right = dfs(root.right);
23         result = Math.max(result, left + right + root.val);
24         return Math.max(0, Math.max(left, right) + root.val);// 不返回小于0的数字！！！
25     }
26 }

 

posted on 2013-09-11 06:34  Step-BY-Step  阅读(199)  评论(0)  编辑  收藏  举报

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