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

GitHub答案源代码请点击这里

Recursive解法 

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        dfs(root, res);
        return res;
    }
    
    private void dfs(TreeNode node, List<Integer> res) {
        if (node == null) {
            return;
        }    
        
        dfs(node.left, res);
        dfs(node.right, res);
        res.add(node.val);
    }
}

Reversed Preorder解法 

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        if (root == null) {
            return new ArrayList<>();
        }
        
        List<Integer> res = new LinkedList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        stack.push(root);
        TreeNode cur = root;
        
        while (!stack.isEmpty()) {
            cur = stack.pop();
            res.add(0, cur.val);
            if (cur.left != null) {
                stack.push(cur.left);
            }
            if (cur.right != null) {
                stack.push(cur.right);
            }
        }
        
        return res;
    }
}

Iterative解法 

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        if (root == null) {
            return new ArrayList<>();
        }
        
        List<Integer> res = new ArrayList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        stack.push(root);
        TreeNode prev = null;
        
        while (!stack.isEmpty()) {
            TreeNode cur = stack.peek();
            if (prev == null || prev.left == cur || prev.right == cur) {
                if (cur.left != null) {
                    stack.push(cur.left);
                } else if (cur.right != null) {
                    stack.push(cur.right);
                } else {
                    stack.pop();
                    res.add(cur.val);
                }
            } else if (prev == cur.left) {
                if (cur.right != null) {
                    stack.push(cur.right);
                } else {
                    stack.pop();
                    res.add(cur.val);
                }
            } else {
                stack.pop();
                res.add(cur.val);
            }
            
            prev = cur;
        }
        
        return res;
    }
}