112. Path Sum

题目：

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

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

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

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

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Tree Depth-first Search

链接：http://leetcode.com/problems/path-sum/

一刷，DFS

class Solution(object):
    def hasPathSum(self, root, sum):
        if not root:
            return False
        if not root.left and not root.right and sum == root.val:
            return True
        return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)

2/17/2017, Java, performance only bit 2%...

用类似前序遍历的方法。一定要注意什么时候需要pop，每次入栈的都是什么值。

注意叶节点是在stack的peek上，空节点不在。

recursive方法远好于iterative算法

 1 public class Solution {
 2     public boolean hasPathSum(TreeNode root, int sum) {
 3         if (root == null) return false;
 4         TreeNode node = root;
 5         TreeNode previous = null;
 6         Stack<TreeNode> stack = new Stack<>();
 7         int value = sum;
 8         stack.push(root);
 9         value -= root.val;
10 
11         while (!stack.isEmpty()) {
12             node = stack.peek();
13             if (previous == null || previous.left == node || previous.right == node) {
14                 if (node.left == null && node.right == null) {
15                      if (value == 0) return true;
16                      else {
17                          previous = stack.pop();
18                          value += previous.val;
19                      }
20                 } else if (node.left != null) {
21                     stack.push(node.left);
22                     value -= node.left.val;
23                     previous = node;
24                 } else if (node.right != null) {
25                     stack.push(node.right);
26                     value -= node.right.val;
27                     previous = node;
28                 }
29             } else if (previous == node.left) {
30                 if (node.right != null) {
31                     stack.push(node.right);
32                     value -= node.right.val;
33                     previous = node;
34                 } else {
35                     previous = stack.pop();
36                     value += previous.val;
37                 }
38 
39             } else if (previous == node.right) {
40                 previous = stack.pop();
41                 value += previous.val;
42             }
43         }
44         return false;
45     }
46 }

5/8/2017

算法班

1 public class Solution {
2     public boolean hasPathSum(TreeNode root, int sum) {
3         if (root == null) return false;
4         if (root.left == null && root.right == null && root.val == sum) return true;
5         return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
6     }
7 }