Leetcode刷题之旅(每日一题)--112. 路径总和

题目描述：

思路：首先针对树第一想法指定是递归。向左右子节点调用sum-当前节点的val。

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

运算结果：

其他做法就是官方提供的广度优先算法：

附上官方代码：

class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        Queue<TreeNode> queNode = new LinkedList<TreeNode>();
        Queue<Integer> queVal = new LinkedList<Integer>();
        queNode.offer(root);
        queVal.offer(root.val);
        while (!queNode.isEmpty()) {
            TreeNode now = queNode.poll();
            int temp = queVal.poll();
            if (now.left == null && now.right == null) {
                if (temp == sum) {
                    return true;
                }
                continue;
            }
            if (now.left != null) {
                queNode.offer(now.left);
                queVal.offer(now.left.val + temp);
            }
            if (now.right != null) {
                queNode.offer(now.right);
                queVal.offer(now.right.val + temp);
            }
        }
        return false;
    }