categories: [Blog,Algorithm]

112. 路径总和

难度简单523
给你二叉树的根节点 root 和一个表示目标和的整数 targetSum ，判断该树中是否存在 根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和 targetSum 。
叶子节点 是指没有子节点的节点。

示例 1：
112. 路径总和 - 图1
输入：root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
输出：true

广度优先

  1. public boolean hasPathSum(TreeNode root, int sum) {
  2.         if (root == null) {
  3.             return false;
  4.         }
  5.         Queue<TreeNode> queNode = new LinkedList<TreeNode>();
  6.         Queue<Integer> queVal = new LinkedList<Integer>();
  7.         queNode.offer(root);
  8.         queVal.offer(root.val);
  9.         while (!queNode.isEmpty()) {
  10.             TreeNode now = queNode.poll();
  11.             int temp = queVal.poll();
  12.             if (now.left == null && now.right == null) {
  13.                 if (temp == sum) {
  14.                     return true;
  15.                 }
  16.                 continue;
  17.             }
  18.             if (now.left != null) {
  19.                 queNode.offer(now.left);
  20.                 queVal.offer(now.left.val + temp);
  21.             }
  22.             if (now.right != null) {
  23.                 queNode.offer(now.right);
  24.                 queVal.offer(now.right.val + temp);
  25.             }
  26.         }
  27.         return false;
  28.     }
  30. 作者：LeetCode-Solution
  31. 链接：https://leetcode-cn.com/problems/path-sum/solution/lu-jing-zong-he-by-leetcode-solution/

递归

  1. /**
  2.  * Definition for a binary tree node.
  3.  * public class TreeNode {
  4.  *     int val;
  5.  *     TreeNode left;
  6.  *     TreeNode right;
  7.  *     TreeNode() {}
  8.  *     TreeNode(int val) { this.val = val; }
  9.  *     TreeNode(int val, TreeNode left, TreeNode right) {
  10.  *         this.val = val;
  11.  *         this.left = left;
  12.  *         this.right = right;
  13.  *     }
  14.  * }
  15.  */
  16. class Solution {
  17.     public boolean hasPathSum(TreeNode root, int sum) {
  18.         if (root == null) {
  19.             return false;
  20.         }
  21.         if (root.left == null && root.right == null) {
  22.             return sum == root.val;
  23.         }
  24.         return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
  25.     }
  27. // 作者：LeetCode-Solution
  28. // 链接：https://leetcode-cn.com/problems/path-sum/solution/lu-jing-zong-he-by-leetcode-solution/
  29. }