Leetcode 226. Invert Binary Tree

最新推荐文章于 2024-10-29 00:15:00 发布

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#tree #dfs #bfs #recursion #iteration

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本文介绍了两种翻转二叉树的方法：深度优先搜索（DFS）的bottom-up预序递归实现和广度优先搜索（BFS）的迭代实现。这两种方法的时间复杂度均为O(n)，但空间复杂度不同，DFS为O(h)，BFS为O(n)。代码分别使用递归和队列来交换二叉树节点的左右子节点，达到翻转效果。

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方法1: dfs-bottomUp-preorder-recursion。时间复杂n，空间复杂h。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root == null) return root;
        TreeNode temp = root.left;
        root.left = root.right == null ? null : invertTree(root.right);
        root.right = temp == null ? null : invertTree(temp);
        return root;
    }
}

方法2: bfs-topDown-iteration-one queue。时间复杂n，空间复杂n。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root == null) return null;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()){
            TreeNode curr = queue.poll();
            TreeNode temp = curr.left;
            curr.left = curr.right;
            curr.right = temp;
            if(curr.left != null) queue.offer(curr.left);
            if(curr.right != null) queue.offer(curr.right);
        }
        return root;
    }
}