leetcode--Symmetric Tree

1.题目描述

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).   For example, this binary tree is symmetric:   1 / \ 2 2 / \ / \ 3 4 4 3 But the following is not:   1 / \ 2 2 \ \ 3 3 Note: Bonus points if you could solve it both recursively and iteratively.

2.解法分析

这个题目其实可以看做是深度搜索的变种，深度搜索有三种：先序、中序和后序，对于这个题目，我们对root的左右子树同时进行先序深度搜索，所不同的是，左子树的深度搜索是“左右”顺序，右子树是“右左”顺序，只要直到深度搜索完成都满足同步，那么这棵树满足要求。

/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isSymmetric(TreeNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function if(!root)return true; vector<TreeNode *>left_traversal; vector<TreeNode *>right_traversal; TreeNode * cur_left = root->left; TreeNode *cur_right = root->right; if((cur_left!=NULL&&cur_right==NULL)||(cur_left==NULL&&cur_right!=NULL))return false; if(cur_left==NULL&&cur_right==NULL)return true; while(!left_traversal.empty()||cur_left) { while(cur_left) { if(!cur_right)return false; if(cur_left->val!=cur_right->val)return false; left_traversal.push_back(cur_left); right_traversal.push_back(cur_right); cur_left= cur_left->left; cur_right = cur_right->right; } if(!left_traversal.empty()) { if(cur_right)return false; cur_left=left_traversal.back();left_traversal.pop_back(); cur_right=right_traversal.back();right_traversal.pop_back(); cur_left = cur_left->right; cur_right = cur_right->left; } }l if(!left_traversal.empty()||!right_traversal.empty())return false; if(cur_left&&cur_right)return cur_left->val==cur_right->val; else { if(!cur_left&&!cur_right)return true; else return false; } return true; } };

转载于:https://www.cnblogs.com/obama/p/3260869.html