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LeetCode-145-二叉树的后序遍历

给定一个二叉树,返回它的 后序 遍历。

相关链接:

  1. {% post_link LeetCode-94-二叉树的中序遍历 %}
  2. {% post_link LeetCode-144-二叉树的前序遍历%}

示例 1:

输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [3,2,1]

解题思路

二叉树的遍历问题都有2种解法,一种是递归,一种是迭代

递归:开启左子树递归,开启右子树递归,添加根节点

迭代:后序遍历的方式是左右根,前序遍历是根左右,如果用Stack来实现根左右,那么左边先加入就会后出,右边后加入会先出,于是看似是add(left)之后add(right),实际上会先访问到right再访问left,从而实现前序遍历得到根右左,即后序遍历的倒序,之后将列表倒序就是后序遍历的结果

迭代模拟:严格按照后序遍历的左右根访问,代码整体与中序遍历很相似,但需要注意其中两个点。

思想来源于这里

第一,stack.peek()只是取出栈顶元素,要和stack.pop()弹出栈顶元素区分开来;

第二,变量last用于保存当前栈顶所弹出的元素,判断 curr.right == last 是为了避免重复访问同一个元素而陷入死循环当中

Java代码(递归)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<Integer> res = new ArrayList<>();
    public List<Integer> postorderTraversal(TreeNode root) {
        if(root==null) return res;
        helper(root);
        return res;
    }
    public void helper(TreeNode root){
        if(root==null) return;
        helper(root.left);
        helper(root.right);
        res.add(root.val);
    }
}

Java代码(迭代Stack)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        LinkedList<Integer> res = new LinkedList<>();
        Stack<TreeNode> stack = new Stack<>(); // Stack实现根右左
        if(root==null) return res;
        stack.add(root);
        while(!stack.isEmpty()){
            TreeNode temp = stack.pop();
            res.addFirst(temp.val); // 添加到头部,实现倒序
            if(temp.left!=null)
                stack.add(temp.left);
            if(temp.right!=null)
                stack.add(temp.right);
        }
        return res;
    }
}

Java代码(迭代模拟)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        Stack<TreeNode> stack = new Stack<>();
        TreeNode curr = root;
        TreeNode last = null;
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            curr = stack.peek();
            if (curr.right == null || curr.right == last) {
                res.add(curr.val);
                stack.pop();
                last = curr;
                curr = null;
            } else {
                curr = curr.right;
            }
        }
        return res;
    }
}



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