Microsoft leetcode (Symmetric Tree)

Symmetric Tree

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

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   /   2   2
 / \ / 3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   /   2   2
   \      3    3

Recursive://1. For every level, if root‘s left == root‘s right then we can say it is symmetric.//2. If root1 && root2 are null return true, if root1 || root2 is null return false.class TreeNode {　　int val;　　TreeNode left;　　TreeNode right;　　public TreeNode(int val) {　　　　this.val = val;　　}}public boolean isSymmetric(TreeNode root) {　　return isMirror(root, root);}public boolean isMirror(TreeNode root1, TreeNode root2) {　　if (root1 == null && root2 == null) return true;　　if (root1 == null || root2 == null) return false;　　return root1.val == root2.val && isMirror(root1.left, root2.right)　　　　&& isMirror(root1.right, root2.left);}

Iterative:DFS -> Stack BFS -> QueueUse stack to traverse all the nodes in the tree and check if left equals to right.

public boolean isSymmetric(TreeNode root) {　　if (root == null) return true;　　Stack<TreeNode> stack = new Stack<>();　　stack.push(root.left);　　stack.push(root.right);　　while(!stack.isEmpty()) {　　　　TreeNode n1 = stack.pop(), n2 = stack.pop();　　　　if (n1 == null && n2 == null) continue;　　　　if (n1 == null || n2 == null || n1.val != n2.val) return false;　　　　else {　　　　　　stack.push(n1.left);　　　　　　stack.push(n2.right);　　　　　　stack.push(n1.right);　　　　　　stack.push(n2.left);　　　　}　　}　　return true;}

原文地址：https://www.cnblogs.com/jjjiajia/p/8450041.html