LeetCode--Symmetric Tree

Problem:

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. confused what “{1,#,2,3}” means? > read more on how

binary tree is serialized on OJ. Subscribe to see which companies

asked this question

Analysis:

方法1：再次创建一个新的函数，该函数有两个参数，一个是左节点，另一个是右节点，利用这两个记录的节点，分别判断左节点的左孩子？=右节点的右孩子；左节点的右孩子？=右节点的左孩子，将判断的结果返回；

这里其实的对两棵子树同时进行递归判断。

方法2：利用堆栈将节点按照特点的顺序（左节点压左孩子，右节点就得压右孩子；左节点压右孩子，右节点就得压左孩子）压入堆栈，把相应的节点作比较，不相同则返回false，相同的话就continue或者压入堆栈。

Anwser1:

递归：

public class Solution {
public boolean isSymmetric(TreeNode root) {
//recursively
if(root==null) return true;
else return isSymmetric(root.left,root.right);
}
public boolean isSymmetric(TreeNode leftNode,TreeNode rightNode){
if(leftNode==null && rightNode==null) return true;
else if (leftNode==null || rightNode==null) return false;
else {
if (leftNode.val==rightNode.val) {
boolean leftSym = isSymmetric(leftNode.left,rightNode.right);
boolean rightSym = isSymmetric(leftNode.right,rightNode.left);
return leftSym && rightSym;
}
else return false;
}
}
}

Anwser2:

迭代

public class Solution {
public boolean isSymmetric(TreeNode root) {
if(root==null) return true;
Stack<TreeNode> leftStack = new Stack<TreeNode>();
Stack<TreeNode> rightStack = new Stack<TreeNode>();
leftStack.push(root.left);
rightStack.push(root.right);
while(leftStack.size()>0 && rightStack.size()>0){
TreeNode leftNode = leftStack.pop();
TreeNode rightNode = rightStack.pop();
if(leftNode==null && rightNode==null) continue;//不能返回true,因为还有其他没有比较的节点。
else if(leftNode==null || rightNode==null) return false;
else if(leftNode.val==rightNode.val){
//压入堆栈的顺序要注意
leftStack.push(leftNode.left);
leftStack.push(leftNode.right);
rightStack.push(rightNode.right);
rightStack.push(rightNode.left);
}
else return false;//考虑的是leftNode.val!=rightNode.val的情况，不能丢
}
return true;//考虑的是完全对称的树，最后应该返回true,不能丢
}
}