[leetcode] 226. Invert Binary Tree 解题报告

本题目可以采用三种方法Straightforward DFS recursive, iterative, BFS solutions

1.深度优先 递归

2.迭代

3.广度优先

a.要弄懂final关键字的含义

此题不加也行

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

 b.迭代

public class Solution {
    public TreeNode invertTree(TreeNode root) {
        
        if (root == null) {
            return null;
        }

        final Deque<TreeNode> stack = new LinkedList<>();
        stack.push(root);
        
        while(!stack.isEmpty()) {
            final TreeNode node = stack.pop();
            final TreeNode left = node.left;
            node.left = node.right;
            node.right = left;
            
            if(node.left != null) {
                stack.push(node.left);
            }
            if(node.right != null) {
                stack.push(node.right);
            }
        }
        return root;
    }
}

c.广度优先

public class Solution {
    public TreeNode invertTree(TreeNode root) {
        
        if (root == null) {
            return null;
        }

        final Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while(!queue.isEmpty()) {
            final TreeNode node = queue.poll();
            final TreeNode left = node.left;
            node.left = node.right;
            node.right = left;

            if(node.left != null) {
                queue.offer(node.left);
            }
            if(node.right != null) {
                queue.offer(node.right);
            }
        }
        return root;
    }
}