【LeetCode】112.Path Sum(Easy)解题报告

【LeetCode】112.Path Sum(Easy)解题报告

题目地址：https://leetcode.com/problems/path-sum/description/

题目描述：

  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.

Solution1:

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
time : O(n)
space : O(n)
递归和非递归两种写法
*/
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if(root == null) return false;
if(root.left == null && root.right == null){
return sum==root.val;
}
return hasPathSum(root.left,sum-root.val) || hasPathSum(root.right,sum-root.val);
}
}

Solution2:

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
time : O(n)
space : O(n)
递归和非递归两种写法
stack
*/
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if(root == null) return false;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode cur = stack.pop();
if(cur.left == null && cur.right == null){
if(cur.val == sum){
return true;
}
}
if(cur.right != null){
stack.push(cur.right);
cur.right.val += cur.val;
}
if(cur.left != null){
stack.push(cur.left);
cur.left.val += cur.val;
}
}
return false;
}
}

Date：2018年3月12日