Problem: 1022. 从根到叶的二进制数之和

题目描述

思路

遍历思想(利用二叉树的先序遍历)

1.在先序遍历的过程中，用一个变量path记录并更新其经过的路径上的值，当遇到根节点时再将其加到结果值res上；
2.该题中要注意数值的二进制操作，该题中path = path << 1 | root.val;即通过不断地左移（由于题目所说数值均为32位整数）并再与当前节点值相或来更新path记录地值

复杂度

时间复杂度:

$O(n)$;其中$n$为二叉树的节点个数

空间复杂度:

$O(h)$；其中$h$为二叉树的高度

Code

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public int sumRootToLeaf(TreeNode root) {traverse(root);return res;}int path = 0;int res = 0;private void traverse(TreeNode root) {if (root == null) {return;}// Leaf rootif (root.left == null && root.right == null) {// Add the value on the path to the resres += path << 1 | root.val;return;}// Change the value of pathpath = path << 1 | root.val;traverse(root.left);traverse(root.right);path = path >> 1;}
}