leetCode解题报告之Binary Tree Postorder Traversal

题目：

Binary Tree Postorder Traversal

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?

分析：

初看以为是很简单的递归后序遍历二叉树的结点，之后看到要返回一个ArrayList之后，感觉如果用平常的递归遍历访问二叉树结点的话，有可能会TLE或者是StackOverFlow，之后想到要用“非递归方式”后序遍历二叉树.

解题思路：

开一个Stack，由于root结点是最后访问的结点，所以我们先将root结点push到Stack中

我们写一个while循环，while循环结束的条件是栈中没有任何结点。

当栈中有结点的时候，我们将取出栈顶结点

1、判断下它是否是叶子结点（left和right都为null）, 如果是叶子结点的话，那么不好意思，把它弹出栈，并把值add到ArrayList中，如果不是叶子结点的话，那么

2、我们判断下它的left（node.left）是否为null,如果不为null，把它的左孩子结点push到栈中来，并把它的左孩子域设为null, 然后跳过此次循环剩下的部分

3、如果它的left 为null, 把它的右孩子结点push到栈中来，并把它的右孩子域设为null，然后跳过此次循环剩下的部分！

图解：

package cn.xym.leetcode.binarytree;

import java.util.ArrayList;
import java.util.Stack;


class TreeNode {
	int val;
	TreeNode left;
	TreeNode right;

	TreeNode(int x) {
		val = x;
	}
}
public class Solution {
	
	public ArrayList<Integer> postorderTraversal(TreeNode root) {
        if (root == null)
        	return null;
        ArrayList<Integer> list = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        //先把最后访问的结点先放入到栈中，即根节点root
        stack.push(root);
        while (stack.size() != 0){
        	TreeNode top = stack.peek();
        	if (top.left == null && top.right == null){
        		list.add(top.val);
        		stack.pop();
        	}
        	if (top.left != null){
        		stack.push(top.left);
        		top.left = null;
        		continue;
        	}
        	if (top.right != null){
        		stack.push(top.right);
        		top.right = null;
        		continue;
        	}
        }
        return list;
    }
}

类似的题目：

Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes‘ values.

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

   1
         2
    /
   3

return [1,2,3].

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

AC代码：

package cn.xym.leetcode.binarytree;

import java.util.ArrayList;
import java.util.Stack;


class TreeNode {
	int val;
	TreeNode left;
	TreeNode right;

	TreeNode(int x) {
		val = x;
	}
}
public class Solution {
	
	public ArrayList<Integer> preorderTraversal(TreeNode root) {
		
		ArrayList<Integer> list = new ArrayList<Integer>();
		ArrayList<Integer> list2 = new ArrayList<Integer>();
        if (root == null)
        	return list;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while (stack.size() != 0){
        	TreeNode top = stack.peek();
        	if (top.left == null && top.right == null){
        		list.add(top.val);
        		stack.pop();
        	}
        	if (top.right != null){
        		stack.push(top.right);
        		top.right = null;
        		continue;
        	}
        	if (top.left != null){
        		stack.push(top.left);
        		top.left = null;
        		continue;
        	}
        }
        for (int i=0; i<list.size(); ++i){
        	list2.add(list.get(list.size()-1-i));
        }
        return list2;
    }
	public static void main(String[] args) {
		Solution sl = new Solution();
		TreeNode root1 = new TreeNode(1);
		TreeNode root2 = new TreeNode(2);
		TreeNode root3 = new TreeNode(3);
		TreeNode root4 = new TreeNode(4);
		TreeNode root5 = new TreeNode(5);
		TreeNode root6 = new TreeNode(6);
		root1.left = root2;
		root1.right = root3;
		root2.left = root4;
		root2.right = root5;
		root3.left = root6;
		
		ArrayList<Integer> list = sl.preorderTraversal(root1);
		System.out.println(list);
	}
}

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leetCode解题报告之Binary Tree Postorder Traversal

原文：http://blog.csdn.net/ljphhj/article/details/21369053