标签：blog 复杂度 rtt solution off public empty lan nbsp
Invert a binary tree.
4 / 2 7 / \ / 1 3 6 9
to
4 / 7 2 / \ / 9 6 3 1
Trivia:
This problem was inspired by this original tweet by Max Howell:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.
分析：翻转二叉树
思路1：
使用递归的方式，时间复杂度为o(n)，空间复杂度为o(n)
JAVA CODE
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode invertTree(TreeNode root) { if(root == null||(root.left==null&&root.right==null)) return root; if(root.left!=null) invertTree(root.left); if(root.right!=null) invertTree(root.right); TreeNode tempL = root.left; TreeNode tempR = root.right; root.left = tempR; root.right = tempL; return root; } }
思路2：
使用迭代的方式，时间复杂度为o(n)，空间复杂度为o(1)
JAVA CODE
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode invertTree(TreeNode root) { if(root == null) return root; Stack<TreeNode> nodes = new Stack<>(); nodes.push(root); while(!nodes.isEmpty()){ TreeNode cur = nodes.pop(); if(cur.right!=null) nodes.push(cur.right); if(cur.left!=null) nodes.push(cur.left); TreeNode temp = cur.left; cur.left = cur.right; cur.right = temp; } return root; } }
标签：blog 复杂度 rtt solution off public empty lan nbsp
原文地址：http://www.cnblogs.com/baichangfu/p/7476698.html
