力扣-112题-路径之和

地址：​​https://leetcode-cn.com/problems/path-sum/​​

java版

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if( root == null){
            return false;
        }
        
        //如果没有左右节点，那么直接用sum-该节点值，跟0对比
        if(root.left == null && root.right == null){
            return sum -root.val == 0 ;
        }

        return hasPathSum(root.left,sum-root.val) || hasPathSum(root.right,sum-root.val);
    }
}

php 版

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     public $val = null;
 *     public $left = null;
 *     public $right = null;
 *     function __construct($value) { $this->val = $value; }
 * }
 */
class Solution {

    /**
     * @param TreeNode $root
     * @param Integer $sum
     * @return Boolean
     */
    function hasPathSum($root, $sum) {
        if(empty($root)){
            return false;
        }
        //如果没有左右节点，那么就是叶子节点
        if(!isset($root->left) && !isset($root->right)){
            return $sum - $root->val == 0;
        }

        return $this->hasPathSum($root->left,$sum-$root->val) || $this->hasPathSum($root->right,$sum - $root->val);
    }
}

go版本

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func hasPathSum(root *TreeNode, sum int) bool {
    
    if root == nil {
        return false
    }

    if root.Left == nil && root.Right == nil {
        return sum == root.Val
    }

    return hasPathSum(root.Left,sum - root.Val) || hasPathSum(root.Right,sum- root.Val)
}