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Symmetric Tree
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
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But the following is not:
1
/ \
2 2
\ \
3 3
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Note:
Bonus points if you could solve it both recursively and iteratively.
解题思路：
题目要求分别用递归法和迭代法做。
1、递归法。思路挺简单，每次检查一对节点，验证这对节点的值是否相同，并且节点1的左孩子与节点2的右孩子的对称的，并且节点1的右孩子与节点2的左孩子是对称的。否则，返回false。
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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSymmetric(TreeNode* root) {
return isSysmetricHelp(root, root);
}
bool isSysmetricHelp(TreeNode* root1, TreeNode* root2){
if(root1==NULL && root2==NULL){
return true;
}
if(root1==NULL || root2==NULL){
return false;
}
if(root1->val != root2->val){
return false;
}
return isSysmetricHelp(root1->left, root2->right)&&isSysmetricHelp(root1->right, root2->left);
}
}; 2、迭代法。有句话，递归转化成非递归，无非就是用栈或堆来存储中间状态。我们用两个队列来存储待比较的两个节点即可。
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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSymmetric(TreeNode* root) {
if(root==NULL){
return true;
}
//两个队列的大小逻辑上一定相同
queue
left({root});
queue
right({root}); while(!left.empty()){ TreeNode* leftNode = left.front(); TreeNode* rightNode = right.front(); left.pop(); right.pop(); if(leftNode->val!=rightNode->val){ return false; } if(leftNode->left!=NULL&&rightNode->right!=NULL){ left.push(leftNode->left); right.push(rightNode->right); }else if(leftNode->left!=NULL || rightNode->right!=NULL){ return false; } if(leftNode->right!=NULL&&rightNode->left!=NULL){ left.push(rightNode->left); right.push(leftNode->right); }else if(leftNode->right!=NULL || rightNode->left!=NULL){ return false; } } return true; } };
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