力扣101.对称二叉树（java)

题目来源

代码1(递归)

java 复制代码

/**
 * 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 boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        return isMirror(root.left,root.right);
    }
    //判断两是否对称
    private boolean isMirror(TreeNode left, TreeNode right) {
        if(left==null && right==null) return true;
        //只有一个为空
        if(left==null || right==null) return false;
        if(left.val != right.val) return false;
        // 左左与右右 左右与右左
        return isMirror(left.left,right.right) && isMirror(left.right,right.left);
    }
}

代码分析

两子树是否对称最小规模子问题就是：

左子树与右子树是否堆成

（对称：左子树与右子树是否都为空，两值是否都为空，不对称：只有一值为空）

左子树的左子树与右子树的右子树是否同时与左子树的右子树与右子树的左子树均为对称

代码2（dfs迭代法）

java 复制代码

/**
 * 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 boolean isSymmetric(TreeNode root) {
        if(root == null) return true;

        Stack<TreeNode> stack = new Stack<>();
        stack.push(root.right);
        stack.push(root.left);

        while(!stack.isEmpty()) {
            TreeNode node1 = stack.pop(); //左子树根结点
            TreeNode node2 = stack.pop(); //右

            if(node1==null && node2==null) continue;
            if(node1==null || node2==null || node1.val!=node2.val) return false;
            // if(node1==null || node2==null) return false;
            // if( node1.val!=node2.val) return false;
            //将......逆向入栈 ： 左左与右右  左右与右左
            // 入栈就是为了 看两子树是否对称
            stack.push(node2.left);
            stack.push(node1.right);
            stack.push(node2.right);
            stack.push(node1.left);
        }
        return true;
    }
}

代码分析

这里的dfs，是双向深入的dfs。因为需要两子树进行比较。

代码3（bfs迭代法）

java 复制代码

/**
 * 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 boolean isSymmetric(TreeNode root) {
        if(root == null) return true;

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

        while(!queue.isEmpty()) {
            TreeNode node1 = queue.poll();
            TreeNode node2 = queue.poll();

            if(node1==null && node2==null) continue;
            if(node1==null || node2==null || node1.val!=node2.val) return false;

            queue.offer(node1.left);
            queue.offer(node2.right);
            queue.offer(node1.right);
            queue.offer(node2.left);
        }
        return true;
    }
}

代码分析

类似dfs，只是逐层处理。