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  • 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.

    time: O(n), space: O(height)

    /**
     * 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 == 0;
            }
            return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
        }
    }
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/10198680.html
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