【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