145. 二叉树的后序遍历

题目描述

给定一个二叉树，返回它的后序遍历。

示例:

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

输出: [3,2,1]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

解法

1. 递归遍历

先递归左右子树，再访问根节点。

2. 栈实现非递归遍历

非递归的思路如下：

先序遍历的顺序是：头、左、右，如果我们改变左右孩子的顺序，就能将顺序变成：头、右、左。

我们先不打印头节点，而是存放到另一个收集栈 ans 中，最后遍历结束，输出收集栈元素，即是后序遍历：左、右、头。收集栈是为了实现结果列表的逆序。我们也可以直接使用链表，每次插入元素时，放在头部，最后直接返回链表即可，无需进行逆序。

3. Morris 实现后序遍历

Morris 遍历无需使用栈，空间复杂度为 O(1)。核心思想是：

遍历二叉树节点，

若当前节点 root 的右子树为空，将当前节点值添加至结果列表 ans 中，并将当前节点更新为 root.left
若当前节点 root 的右子树不为空，找到右子树的最左节点 next（也即是 root 节点在中序遍历下的后继节点）：
- 若后继节点 next 的左子树为空，将当前节点值添加至结果列表 ans 中，然后将后继节点的左子树指向当前节点 root，并将当前节点更新为 root.right。
- 若后继节点 next 的左子树不为空，将后继节点左子树指向空（即解除 next 与 root 的指向关系），并将当前节点更新为 root.left。
循环以上步骤，直至二叉树节点为空，遍历结束。
最后返回结果列表的逆序即可。

Morris 后序遍历跟 Morris 前序遍历思路一致，只是将前序的“根左右”变为“根右左”，最后逆序结果即可变成“左右根”。

Python3

递归：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        def dfs(root):
            if root is None:
                return
            dfs(root.left)
            dfs(root.right)
            nonlocal ans
            ans.append(root.val)

        ans = []
        dfs(root)
        return ans

栈实现非递归：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        ans = []
        if root is None:
            return ans
        stk = [root]
        while stk:
            node = stk.pop()
            ans.append(node.val)
            if node.left:
                stk.append(node.left)
            if node.right:
                stk.append(node.right)
        return ans[::-1]

Morris 遍历：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        ans = []
        while root:
            if root.right is None:
                ans.append(root.val)
                root = root.left
            else:
                next = root.right
                while next.left and next.left != root:
                    next = next.left
                if next.left != root:
                    ans.append(root.val)
                    next.left = root
                    root = root.right
                else:
                    next.left = None
                    root = root.left
        return ans[::-1]

Java

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 {
    private List<Integer> ans;

    public List<Integer> postorderTraversal(TreeNode root) {
        ans = new ArrayList<>();
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.left);
        dfs(root.right);
        ans.add(root.val);
    }
}

栈实现非递归：

/**
 * 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 {
    public List<Integer> postorderTraversal(TreeNode root) {
        LinkedList<Integer> ans = new LinkedList<>();
        if (root == null) {
            return ans;
        }
        Deque<TreeNode> stk = new ArrayDeque<>();
        stk.push(root);
        while (!stk.isEmpty()) {
            TreeNode node = stk.pop();
            ans.addFirst(node.val);
            if (node.left != null) {
                stk.push(node.left);
            }
            if (node.right != null) {
                stk.push(node.right);
            }
        }
        return ans;
    }
}

Morris 遍历：

/**
 * 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) {
        LinkedList<Integer> ans = new LinkedList<>();
        while (root != null) {
            if (root.right == null) {
                ans.addFirst(root.val);
                root = root.left;
            } else {
                TreeNode next = root.right;
                while (next.left != null && next.left != root) {
                    next = next.left;
                }
                if (next.left == null) {
                    ans.addFirst(root.val);
                    next.left = root;
                    root = root.right;
                } else {
                    next.left = null;
                    root = root.left;
                }
            }
        }
        return ans;
    }
}

TypeScript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function postorderTraversal(root: TreeNode | null): number[] {
    if (root == null) return [];
    let stack = [];
    let ans = [];
    let prev = null;
    while (root || stack.length) {
        while (root) {
            stack.push(root);
            root = root.left;
        }
        root = stack.pop();
        if (!root.right || root.right == prev) {
            ans.push(root.val);
            prev = root;
            root = null;
        } else {
            stack.push(root);
            root = root.right;
        }
    }
    return ans;
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> ans;
        while (root)
        {
            if (!root->right)
            {
                ans.push_back(root->val);
                root = root->left;
            }
            else
            {
                TreeNode* next = root->right;
                while (next->left && next->left != root)
                {
                    next = next->left;
                }
                if (!next->left)
                {
                    ans.push_back(root->val);
                    next->left = root;
                    root = root->right;
                }
                else
                {
                    next->left = nullptr;
                    root = root->left;
                }
            }
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func postorderTraversal(root *TreeNode) []int {
	var ans []int
	for root != nil {
		if root.Right == nil {
			ans = append([]int{root.Val}, ans...)
			root = root.Left
		} else {
			next := root.Right
			for next.Left != nil && next.Left != root {
				next = next.Left
			}
			if next.Left == nil {
				ans = append([]int{root.Val}, ans...)
				next.Left = root
				root = root.Right
			} else {
				next.Left = nil
				root = root.Left
			}
		}
	}
	return ans
}

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145. 二叉树的后序遍历

题目描述

解法

Python3

Java

TypeScript

C++

Go

...

简介

发行版

贡献者

近期动态

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145. 二叉树的后序遍历

题目描述

解法

Python3

Java

TypeScript

C++

Go

...

简介

发行版

贡献者

近期动态

搜索帮助

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