leetcode 101. 对称二叉树 #树

[webVueBlog]

101. 对称二叉树

[webVueBlog]

101. 对称二叉树

/*
 * @lc app=leetcode.cn id=101 lang=javascript
 *
 * [101] 对称二叉树
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
进阶：你可以运用递归和迭代两种方法解决这个问题吗？
(64 ms)
 */
// 迭代
var isSymmetric = function(root) {
 let queue = [root, root];
 while(queue.length > 0) {
  let r1 = queue.shift();
  let r2 = queue.shift();
  if(!r1 && !r2) continue;
  if(!r1 || !r2) return false;
  if(r1.val !== r2.val) return false;
  queue.push(r1.left)
  queue.push(r2.right)
  queue.push(r1.right)
  queue.push(r2.left);
 }
 return true
}

//递归
// var isSymmetric = function(root) {
//  function isMirror(r1, r2) {
//   if(!r1 && !r2) return true
//   if(!r1 || !r2) return false
//   return r1.val === r2.val && isMirror(r1.left, r2.right) && isMirror(r1.right, r2.left)
//  }
//  return isMirror(root, root)
// }

//  var isSymmetric = function(root) {
//   if (!root) return true;
  
//   return symmetryChecker(root.left, root.right);
// };

// function symmetryChecker(left, right) {
//   if (left == null && right == null) return true;
//   if (left == null || right == null) return false;
//   if (left.val !== right.val) return false;
  
//   return symmetryChecker(left.left, right.right) && symmetryChecker(left.right, right.left);
// }
// @lc code=end