112. Path Sum

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.

一刷
题解：这题最容易出错的部分是要求从root到leaf存在一个path, 从而判断是否是leaf的方式：root.left == null && root.right == null && sum == root.val

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
        if(root.left == null && root.right == null && sum == root.val) return true;
        return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum - root.val);
       
    }
}

二刷
思路同上

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
        if(root.left == null && root.right == null && sum == root.val) return true;
        return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum - root.val);
    }
}

三刷
recursion

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
        if(root.val == sum && root.left == null && root.right == null) return true;
        return hasPathSum(root.left, sum-root.val) || 
            hasPathSum(root.right, sum-root.val);
    }
}