一天一大 lee(翻转二叉树)难度:简单-Day20200916

题目:[1]

翻转一棵二叉树。

示例：

     4
   /   
  2     7
 /    / 
1   3 6   9

输出

     4
   /   
  7     2
 /    / 
9   6 3   1

抛砖引玉

思路

深度优先遍历（DFS）

传入的节点如果不是叶子节点即包含左右子树，则优先翻转左右子树
递归到叶子节点开始翻转，之后向上逐层交换拼接

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var invertTree = function(root) {
  if (root === null) {
    return null
  }
  let left = invertTree(root.left),
    right = invertTree(root.right)
  root.left = right
  root.right = left
  return root
}

广度优先遍历（BFS）

从根结点到叶子节点逐层入栈
从栈内逐个取出翻转，遇到非叶子节点继续入栈

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var invertTree = function(root) {
  let queue = [root]
  while (queue.length) {
    let node = queue.shift()
    if (node) {
      let left = node.left
      node.left = node.right
      node.right = left
      queue.push(node.left)
      queue.push(node.right)
    }
  }

  return root
}