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Path Sum leetcode java

2024-07-14 23:38:03   224人阅读

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.

 

题解：

还是对树的操作，递归的解法：

 1     public boolean hasPathSum(TreeNode root, int sum) {
 2         if(root == null) 
 3             return false;
 4         
 5         sum -= root.val;
 6         if(root.left == null && root.right==null)  
 7             return sum == 0;
 8         else 
 9             return hasPathSum(root.left,sum) || hasPathSum(root.right,sum);
10     }

 

非递归的解法（Reference：http://www.programcreek.com/2013/01/leetcode-path-sum/）：

 

 1     public boolean hasPathSum(TreeNode root, int sum) {
 2         if(root == null) return false;
 3  
 4         LinkedList<TreeNode> nodes = new LinkedList<TreeNode>();
 5         LinkedList<Integer> values = new LinkedList<Integer>();
 6  
 7         nodes.add(root);
 8         values.add(root.val);
 9  
10         while(!nodes.isEmpty()){
11             TreeNode curr = nodes.poll();
12             int sumValue = values.poll();
13  
14             if(curr.left == null && curr.right == null && sumValue=http://www.mamicode.com/=sum){
15                 return true;
16             }
17  
18             if(curr.left != null){
19                 nodes.add(curr.left);
20                 values.add(sumValue+curr.left.val);
21             }
22  
23             if(curr.right != null){
24                 nodes.add(curr.right);
25                 values.add(sumValue+curr.right.val);
26             }
27         }
28  
29         return false;
30     }


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