226. Invert Binary Tree

Leetcode Tree

Invert a binary tree.

Example:

Input:

     4
   /   \
  2     7
 / \   / \
1   3 6   9
Output:

     4
   /   \
  7     2
 / \   / \
9   6 3   1

分析

最开始想到的方法是递归，而且也很简单。

    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        TreeNode tmp = root.right;
        root.right = root.left;
        root.left = tmp;
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }

使用DFS的递归:

class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        TreeNode tmpRight = root.right;
        root.right = invertTree(root.left);
        root.left = invertTree(tmpRight);
        return root;
    }
}

递归的时间和空间复杂度是$O(n)$, $n$是节点数。

还有迭代的方法

public TreeNode invertTreeQueue(TreeNode root) {
    if (root == null) return null;
    Queue<TreeNode> queue = new LinkedList<TreeNode>();
    queue.add(root);
    while (!queue.isEmpty()) {
        TreeNode current = queue.poll();
        TreeNode temp = current.left;
        current.left = current.right;
        current.right = temp;
        if (current.left != null) queue.add(current.left);
        if (current.right != null) queue.add(current.right);
    }
    return root;
}