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?

Ref: http://www.cnblogs.com/feiling/p/3426486.html

[解题思路]

后序遍历的非递归相对来说比较难，根节点需要在其左右孩子都访问结束后才能被访问，因此对于任一节点，先将其入栈，如果p不存在左孩子和右孩子，则可以直接访问它；或者p存在左孩子或者右孩子，但是其左孩子和右孩子都已被访问过了，则同样可以直接访问该节点。若非上述两种情况，则将p的右孩子和左孩子依次入栈，这样就保证了每次去站定元素的时候，左孩子在右孩子前面被访问，左孩子和右孩子在根节点前面被访问。

Iteration

/**
 * 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){
            return result;
        }
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode cur = null, pre = null;
        stack.push(root);
        
        while(!stack.empty()){
            cur = stack.peek();
            // pre 用来记录上一个访问的节点，防止root 重复访问左节点和右节点
            // (cur.left == pre || cur.right == pre) 可能root的右子为空,所以             // 上一个访问的就为左节点
            if((cur.left == null && cur.right == null) ||
               ((pre != null) && (cur.left == pre || cur.right == pre))){
                result.add(cur.val);
                pre = cur;
                stack.pop();
            } else {
                if(cur.right != null){
                    stack.push(cur.right);
                }
                if(cur.left != null){
                    stack.push(cur.left);
                }
            }
            
        }
        
        return result;
    }
}

Recursion

/**
 * 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)
                return result;
            if(root.left != null){
                result.addAll(postorderTraversal(root.left));
            }
            if(root.right != null){
                result.addAll(postorderTraversal(root.right));
            }
            result.add(root.val);
            
        return result;
    }
}

DP

/**
 * 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>();
          
        postorder(root, result);
            
        return result;
    }
    
    private void postorder(TreeNode root, ArrayList<Integer> result){
        if(root != null){
            postorder(root.left, result);
            postorder(root.right, result);
            result.add(root.val);
        }
    }
}