Path Sum

来源：互联网发布：yum wget 编辑：程序博客网时间：2024/06/05 05:22

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.

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean hasPathSum(TreeNode root, int sum) {        if(root == null){            return false;        }        return inorder(root,0,sum);    }    public boolean inorder(TreeNode root ,int value , int sum){//value 是父节点传递下来的值，sum是要求达到的sum值        if(root.left == null && root.right == null){//叶子节点            if(value + root.val == sum){                return true;            }else{                return false;            }        }else{//非叶子节点            boolean left = false;            if(root.left != null){                left = inorder(root.left,root.val+value,sum);            }            boolean right = false;            if(root.right != null){                right = inorder(root.right,root.val+value,sum);            }            return left || right;        }            }}

跟这一道题相比，还是比较简单的，思路很简单，一个DFS，找到叶子时也就找到了这条路径所有节点之和，然后判断即可。

Runtime: 254 ms

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