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 andsum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.

Analysis

递归求解,注意的是对path的定义:root-to-leaf path

因此递归返回的条件之一是

root.val == sum && root.left == null && root.right == null

左右子树有一个满足即可 (给子树的sum是每次传入的sum - root.val)

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

Solution

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

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