leetcode--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?

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        if(root != null){
            Stack<TreeNode> sta = new Stack<TreeNode>();
            sta.push(root);
            HashSet<TreeNode> hset = new HashSet<TreeNode>();
            hset.add(root);
            while(!sta.empty()){
                TreeNode aNode = sta.pop();
                if(aNode.left != null && hset.add(aNode.left)){
                    sta.push(aNode);
                    sta.push(aNode.left);        
                }
                else if(aNode.right != null && hset.add(aNode.right)){
                    sta.push(aNode);
                    sta.push(aNode.right);                
                }
                else
                    result.add(aNode.val);
            } 
        }
        return result;
    }
}

Another solution:

public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
       List<Integer> postOrderTreeValue = new ArrayList<Integer>();
		Stack<TreeNode> TreeNodeStack = new Stack<TreeNode>();
		TreeNode haveDoneRightNode = null; 
		
		//initialize the stack with TreeNodes
		if(root != null)
			putTreeNodeInStack(TreeNodeStack, root);
		while(!TreeNodeStack.isEmpty()){
			TreeNode currentNode = TreeNodeStack.pop();
			if(currentNode.right == null || currentNode.right == haveDoneRightNode){
				postOrderTreeValue.add(currentNode.val);
				if(!TreeNodeStack.isEmpty() && currentNode == TreeNodeStack.peek().right)
					haveDoneRightNode = currentNode;
			}
			else{
				TreeNodeStack.push(currentNode);
				putTreeNodeInStack(TreeNodeStack, currentNode.right);
			}
		}
		return postOrderTreeValue;
	}
	
	private void putTreeNodeInStack(Stack<TreeNode> TreeNodeStack, TreeNode root){
		while(root !=null){
			TreeNodeStack.push(root);
			if(root.left != null)
				root = root.left;
			else if(root.right != null)
				root = root.right;
			else
				root = null;
		}
    }
}

 

原文地址:https://www.cnblogs.com/averillzheng/p/3553030.html