112. Path Sum(js)

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.

Note: A leaf is a node with no children.

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.

题意:是否存在一条从根节点至叶子节点的路径值和与sum相等

代码如下:

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} sum
 * @return {boolean}
 */
var hasPathSum = function(root, sum) {
    if(!root) return false;
    if(root.left===null && root.right===null && root.val===sum) return true;
    return hasPathSum(root.left,sum-root.val) || hasPathSum(root.right,sum-root.val);
};
原文地址:https://www.cnblogs.com/xingguozhiming/p/10747292.html