LeetCode 145 Binary Tree Postorder Traversal (后序遍历二叉树)

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Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1\2/3

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

题目链接：https://leetcode.com/problems/binary-tree-postorder-traversal/

递归：

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

非递归：后序遍历左右中，考虑到栈的先进先出，故先将右子树压栈，两种情况记录答案出栈，1是遇到叶子，2是父节点的子节点已经被访问过，根据压栈顺序，可以保证这样可以完成后序遍历

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
public class Solution {public List<Integer> postorderTraversal(TreeNode root) {List<Integer> ans = new ArrayList<>();Stack<TreeNode> stk = new Stack<>();if(root != null) {stk.push(root);}TreeNode cur = null;TreeNode pre = null;while(!stk.empty()) {cur = stk.peek();if(cur.left == null && cur.right == null) {ans.add(cur.val);pre = cur;stk.pop();}else if(pre != null && (cur.left == pre || cur.right == pre)) {ans.add(cur.val);pre = cur;stk.pop();}else {if(cur.right != null) {stk.push(cur.right);}if(cur.left != null) {stk.push(cur.left);}}}return ans;}
}

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