leetcode--Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and  sum = 22 ,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

题意：给定二叉树和一个sum，找出所有和为sum的路径

分类：二叉树

解法1：后序遍历非递归算法。每次访问一个节点，判断当前和是否和sum相等，如果是则保存这个路径。

[java]  view plain  copy

1. /** 
2.  * Definition for a binary tree node. 
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.     public List<List<Integer>> pathSum(TreeNode root, int target) {  
12.         ArrayList<TreeNode> stack = new ArrayList<TreeNode>();    
13.         List<List<Integer>> result = new ArrayList<List<Integer>>();    
14.         int top = -1;    
15.         int sum = 0;    
16.         TreeNode p = root;          
17.         do{             
18.             while(p!=null){    
19.                 top++;    
20.                 sum += p.val;    
21.                 stack.add(top,p);                   
22.                 p = p.left;                 
23.             }               
24.             boolean flag = true;    
25.             TreeNode q = null;    
26.             while(top!=-1 && flag){    
27.                 p = stack.get(top);    
28.                 if(p.right==q){    
29.                     if(p.left==null&&p.right==null && sum==target){    
30.                         ArrayList<Integer> t = new ArrayList<Integer>();    
31.                         for(TreeNode tNode : stack){    
32.                             t.add(tNode.val);    
33.                         }    
34.                         result.add(t);    
35.                     }    
36.                     sum -= p.val;    
37.                     stack.remove(top);    
38.                     top--;    
39.                     q = p;    
40.                 }else{    
41.                     p = p.right;    
42.                     flag = false;                   
43.                 }    
44.             }    
45.         }while(top!=-1);    
46.         return result;    
47.     }         
48. } 

原文链接http://blog.youkuaiyun.com/xiaosongluo/article/details/52797156