leetcode-二叉树

xkge / 2023-09-06 / 原文

二叉树遍历用递归的方式比较简单，但是迭代还是稍微有点绕，记录一下二叉树迭代遍历的统一框架，以防忘记：

主要的思路依旧是栈解决，但是为了当前栈顶元素是否需要被加入到result list中，巧妙地在需要被加入到result list中的元素之前加上一个null以示区分。

102. 二叉树的层序遍历 - 力扣（LeetCode）

代码如下：

/**
 * 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 List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        Stack<TreeNode> stack = new Stack<>();
        if(root == null) return res;
        stack.push(root);
        while(stack.size() > 0){
            TreeNode cur = stack.pop();
            if(cur != null){
               //先push右节点，因为stack具有FIFO的特性
                if(cur.right != null) stack.push(cur.right);
               //push当前节点，并且用null标记
                stack.push(cur);
                stack.push(null);
                //push左节点
                if(cur.left != null) stack.push(cur.left);
            }
            //如果当前栈顶元素为null，则说明需要加入到list当中
            else{
                res.add(stack.pop().val);
            }
        }
        return res;
    }
}

先序和后序只需要调整左中右三个节点的入栈顺序，非常方便。

226. 翻转二叉树 - 力扣（LeetCode）

可以通过先序遍历或者后序遍历，一边遍历一边反转。（中序遍历不行，会导致多次反转）

/**
 * 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 {
    //先序遍历反转
    public TreeNode invertTree(TreeNode root) {
        dfs(root);
        return root;
    }

    public void dfs(TreeNode node){
        if(node == null) return;
        //反转
        TreeNode temp = node.left;
        node.left = node.right;
        node.right = temp;
        dfs(node.left);
        dfs(node.right);
    }
}

101. 对称二叉树 - 力扣（LeetCode）

判断二叉树是否对成，可以采用迭代法或者递归法，本质就是一个一个节点比较：

/**
 * 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 {
    //迭代法，简单来说就是左右子树一个一个节点进行比较，可以用队列或者栈来存储每个子树的节点
    public boolean isSymmetric(TreeNode root) {
        Stack<TreeNode> stackLeft = new Stack<>();
        Stack<TreeNode> stackRight = new Stack<>();
        stackLeft.push(root.left);
        stackRight.push(root.right);
        while(!stackLeft.isEmpty() && !stackRight.isEmpty()){
            TreeNode left = stackLeft.pop();
            TreeNode right = stackRight.pop();
            if(left == null && right == null) continue;
            if(left == null || right == null) return false;
            if(left.val != right.val) return false;
            //按照比较顺序加入即可
            stackLeft.push(left.left);
            stackLeft.push(left.right);
            stackRight.push(right.right);
            stackRight.push(right.left);

        }
        if(!stackLeft.isEmpty() || !stackRight.isEmpty()) return false;
        return true;
    }
}

//递归法
class Solution {
    public boolean isSymmetric(TreeNode root) {
        return compare(root.left, root.right);
    }

    public boolean compare(TreeNode left, TreeNode right){
        //四种情况
        //两方均为空，返回true
        if(left == null && right == null) return true;
        //一方为空，返回false
        if(left == null || right == null) return false;
        //两方均不为空，但值不相等，返回false
        if(left.val != right.val) return false;

        //以上条件均不满足，则说明left.val == right.val，此时要进一步进行判断
        boolean compareInside = compare(left.right, right.left);
        boolean compareOutside = compare(left.left, right.right);
        return compareInside && compareOutside;
    }
}