Question

145. 二叉树的后序遍历

English Version

题目描述

给你一棵二叉树的根节点 root ，返回其节点值的 后序遍历 。

 

示例 1：

输入：root = [1,null,2,3]
输出：[3,2,1]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1]
输出：[1]

 

提示：

• 树中节点的数目在范围 [0, 100] 内
• -100 <= Node.val <= 100

 

进阶：递归算法很简单，你可以通过迭代算法完成吗？

解法

方法一：递归

我们先递归左右子树，然后再访问根节点。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数，空间复杂度主要取决于递归调用的栈空间。

# 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)
      ans.append(root.val)

    ans = []
    dfs(root)
    return ans

/**
 * 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 = new ArrayList<>();

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

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

/**
 * 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;
    function<void(TreeNode*)> dfs = [&](TreeNode* root) {
      if (!root) {
        return;
      }
      dfs(root->left);
      dfs(root->right);
      ans.push_back(root->val);
    };
    dfs(root);
    return ans;
  }
};

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

/**
 * 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[] {
  const ans: number[] = [];
  const dfs = (root: TreeNode | null) => {
    if (!root) {
      return;
    }
    dfs(root.left);
    dfs(root.right);
    ans.push(root.val);
  };
  dfs(root);
  return ans;
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, ans: &mut Vec<i32>) {
    if root.is_none() {
      return;
    }
    let node = root.as_ref().unwrap().borrow();
    Self::dfs(&node.left, ans);
    Self::dfs(&node.right, ans);
    ans.push(node.val);
  }

  pub fn postorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
    let mut ans = vec![];
    Self::dfs(&root, &mut ans);
    ans
  }
}

方法二：栈实现后序遍历

先序遍历的顺序是：根、左、右，如果我们改变左右孩子的顺序，就能将顺序变成：根、右、左。最后再将结果反转一下，就得到了后序遍历的结果。

因此，栈实现非递归遍历的思路如下：

1. 定义一个栈 $stk$，先将根节点压入栈
2. 若栈不为空，每次从栈中弹出一个节点
3. 处理该节点
4. 先把节点左孩子压入栈，接着把节点右孩子压入栈（如果有孩子节点）
5. 重复 2-4
6. 将结果反转，得到后序遍历的结果

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数，空间复杂度主要取决于栈空间。

# 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]

/**
 * 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;
  }
}

/**
 * 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;
    if (!root) {
      return ans;
    }
    stack<TreeNode*> stk;
    stk.push(root);
    while (stk.size()) {
      auto node = stk.top();
      stk.pop();
      ans.push_back(node->val);
      if (node->left) {
        stk.push(node->left);
      }
      if (node->right) {
        stk.push(node->right);
      }
    }
    reverse(ans.begin(), ans.end());
    return ans;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func postorderTraversal(root *TreeNode) (ans []int) {
  if root == nil {
    return
  }
  stk := []*TreeNode{root}
  for len(stk) > 0 {
    node := stk[len(stk)-1]
    stk = stk[:len(stk)-1]
    ans = append(ans, node.Val)
    if node.Left != nil {
      stk = append(stk, node.Left)
    }
    if node.Right != nil {
      stk = append(stk, node.Right)
    }
  }
  for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
    ans[i], ans[j] = ans[j], ans[i]
  }
  return
}

/**
 * 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[] {
  const ans: number[] = [];
  if (!root) {
    return ans;
  }
  const stk: TreeNode[] = [root];
  while (stk.length) {
    const { left, right, val } = stk.pop();
    ans.push(val);
    left && stk.push(left);
    right && stk.push(right);
  }
  ans.reverse();
  return ans;
}

方法三：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，

1. 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
2. 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。
3. 循环以上步骤，直至二叉树节点为空，遍历结束。
4. 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”。

时间复杂度 $O(n)$，其中 $n$ 是二叉树的节点数。空间复杂度 $O(1)$。

# 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]

/**
 * 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;
  }
}

/**
 * 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 != root) {
          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;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func postorderTraversal(root *TreeNode) (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
}

/**
 * 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[] {
  const ans: number[] = [];
  while (root !== null) {
    const { val, left, right } = root;
    if (right === null) {
      ans.push(val);
      root = left;
    } else {
      let next = right;
      while (next.left !== null && next.left !== root) {
        next = next.left;
      }
      if (next.left === null) {
        ans.push(val);
        next.left = root;
        root = right;
      } else {
        next.left = null;
        root = left;
      }
    }
  }
  return ans.reverse();
}