226. Invert Binary Tree

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本文介绍了三种翻转二叉树的方法：递归深度优先搜索(DFS)、使用栈实现DFS及使用队列进行广度优先搜索(BFS)。每种方法都详细展示了如何交换二叉树节点的左右子节点，最终实现整棵树的翻转。

题目描述：同点击打开链接

Invert a binary tree.

class TreeNode {
      int val;
      TreeNode left;
      TreeNode right;
      TreeNode(int x) { val = x; }
}

思路一：递归DFS

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

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

思路二：用栈，DFS

class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        if (root.left == null && root.right == null) return root;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty())
        {
            TreeNode node = stack.pop();
            if (node.left != null || node.right != null)
            {
                TreeNode tmp = node.left;
                node.left = node.right;
                node.right = tmp;
            }
            if (node.left != null) stack.push(node.left);
            if (node.right != null) stack.push(node.right);
        }
        return root;
    }
}

思路三：用队列，BFS

class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        if (root.left == null && root.right == null) return root;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty())
        {
            TreeNode node = queue.poll();
            if (node.left != null || node.right != null)
            {
                TreeNode tmp = node.left;
                node.left = node.right;
                node.right = tmp;
            }
            if (node.left != null) queue.add(node.left);
            if (node.right != null) queue.add(node.right);
        }
        return root;
    }
}