🌕 145. 二叉树的后序遍历

难度: 🌕

问题描述

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解法

/**
 * 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> postorderTraversal(TreeNode root) {
        // 思路：
        // 迭代 - 借助 辅助栈
        // 先序顺序为 root -> left -> right
        // 后序顺序为 left -> right -> root
        LinkedList<TreeNode> stack = new LinkedList<>();
        LinkedList<Integer> tmpStack = new LinkedList<>();
        if(root == null) {
            return new ArrayList<Integer>();
        }
        // root != null
        stack.push(root);
        while(!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            tmpStack.push(cur.val);
            if(cur.left != null) {
                stack.push(cur.left);
            }
            if(cur.right != null) {
                stack.push(cur.right);
            }
        }
        // tmpStack 中顺序为 后序遍历的反序
        List<Integer> res = new ArrayList<>();
        while(!tmpStack.isEmpty()) {
            res.add(tmpStack.pop());
        }
        return res;
    }
}

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输出

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解法 2 - 递归

/**
 * 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 {
    List<Integer> res = new ArrayList<>();
    public List<Integer> postorderTraversal(TreeNode root) {
        // 思路：
        // 递归
        mySol(root);
        return res;
    }

    private void mySol(TreeNode root) {
        // 递归终止条件
        if(root == null) {
            return;
        }
        // root != null
        mySol(root.left);
        mySol(root.right);
        res.add(root.val);
    }
}

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输出 2