LeetCode题解-145-Binary Tree Postorder Traversal

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原题

解法概览

解法1为递归；

解法2位迭代。

解法1

解题思路

递归法。后序遍历左子树，后序遍历右子树，访问根节点。

代码

public class Solution145_recursive {
public List<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> treeValList = new ArrayList<Integer>();

if (root == null)
return treeValList;

ArrayList<Integer> leftTreeValList = (ArrayList<Integer>) postorderTraversal(root.left);
ArrayList<Integer> rightTreeValList = (ArrayList<Integer>) postorderTraversal(root.right);

treeValList.addAll(leftTreeValList);
treeValList.addAll(rightTreeValList);
treeValList.add(root.val);

return treeValList;
}
}

解法2

解题思路

1)当栈顶指针非空，应遍历左子树（遍历时不访问节点）；
2)若从左子树返回，如果当前节点的右子树为空或者已被访问，那么右子树进栈，否则出栈，访问该节点。

图解

看不清图的话可以在浏览器 页面中单独打开。

代码

<pre name="code" class="java">public class Solution145_iterator {
public List<Integer> postorderTraversal(TreeNode root) {
Stack<TreeNode> treeNodeStack = new Stack<TreeNode>();
ArrayList<Integer> treeNodeList = new ArrayList<Integer>();
treeNodeStack.push(root);
TreeNode lastVisit = null;

while (!treeNodeStack.empty()){
while (treeNodeStack.peek() != null){
TreeNode currentNode = treeNodeStack.peek();
treeNodeStack.push(currentNode.left);
}
treeNodeStack.pop();

if (!treeNodeStack.isEmpty()){
TreeNode currentNode = treeNodeStack.peek();
if (currentNode.right == null || currentNode.right == lastVisit){
lastVisit = treeNodeStack.pop();
treeNodeList.add(lastVisit.val);
treeNodeStack.push(null);
}
else{
treeNodeStack.push(currentNode.right);
}
}
}
return treeNodeList;
}
}