101. Symmetric tree (tree, Queue; DFS, WFS)

Given a binary tree, check whether it is a mirror of the itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1   /   2   2/\/3  4 4  3

But the following are not:

    1   /   2   2   \      3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

Idea: To wait until the left son and the right son of the results are known, can judge the current node symmetry, so is the post-order traversal

Method I: Recursive sequential traversal

classSolution { Public:    BOOLIssymmetric (TreeNode *root) {        if(!root)return true; BOOLresult; if((root->left==null && root->right! = NULL) | | (root->left!=null && Root->right = =NULL)) {            return false; }        Else if(Root->left = = NULL && Root->right = =NULL) {            return true; }        Else{result= CMP (Root->left, root->Right ); }          }    BOOLCMP (TreeNode * Node1, treenode*Node2) {        intRESULT1 =true; intRESULT2 =true; if(Node1->val!=node2->val)return false; Else        {            //Recursive end condition: at least one of the nodes is null            if((node1->left==null && node2->right! = NULL) | |(Node1->left!=null && node2->right = NULL) | |(Node1->right!=null && node2->left = NULL) | |(Node1->right==null && Node2->left! =NULL)) {                return false; }            if((Node1->left = = NULL && Node2->right = = null) &&(Node1->right = = NULL && node2->left==NULL)) {                return true; }            //compare the two points to each other to compare the left son of node 1 and the right son of Node 2, as well as the right son of node 1 and the left son of Node 2            if(Node1->left! =NULL) {RESULT1= CMP (node1->left,node2->Right ); }            if(Node1->right! =NULL) {RESULT2= CMP (node1->right,node2->Left ); }             return(Result1 &&result2); }    }     };

Method II: Hierarchical Traversal with queues (hierarchy traversal is always implemented with queues)

classSolution { Public:    BOOLIssymmetric (TreeNode *root) {        if(Root = NULL)return true; Queue<TreeNode*>Q; Q.push (Root-Left ); Q.push (Root-Right ); TreeNode*T1, *T2;  while(!Q.empty ()) {T1=Q.front ();            Q.pop (); T2=Q.front ();            Q.pop (); if(T1 = = NULL && T2 = =NULL)Continue; if(T1 = = NULL | | t2 = = NULL | | T1->val! = t2->val)return false; Q.push (T1-Left ); Q.push (T2-Right ); Q.push (T1-Right ); Q.push (T2-Left ); }        return true; }};

101. Symmetric tree (tree, Queue; DFS, WFS)