README.md

Path Sum

Description

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.

Tags: Tree, Depth-first Search

思路

题意是查找二叉树中是否存在从根结点到叶子的路径和为某一值，利用深搜在遇到叶子节点时判断是否满足即可。

Java:

/**
 * 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;
        if (root.left == null && root.right == null) return sum == root.val;
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}

kotlin(216ms/100.00%):

class Solution {
    fun hasPathSum(root: TreeNode?, sum: Int): Boolean {
        if (root == null) return false
        if (root.left == null && root.right == null) return sum == root.`val`
        return hasPathSum(root.right, sum - root.`val`) || hasPathSum(root.left, sum - root.`val`)
    }
}

JavaScript：

var hasPathSum = function(root, sum) {
  if(root == null) {
      return false   
  }
  if(root.val === sum  && root.left === null && root.right === null) {
      return true
  }
  let sumNext = sum - root.val
  return hasPathSum(root.left, sumNext) || hasPathSum(root.right, sumNext)
};

结语

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