关于java:二叉树的遍历Java版

递归最简略，迭代要用栈（其实也是模仿递归）。

前序遍历

递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) {this.val = val;}
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {List<Integer> res = new ArrayList<Integer>();
    public List<Integer> preorderTraversal(TreeNode root) {if(root == null) return res;
        res.add(root.val);
        preorderTraversal(root.left);
        preorderTraversal(root.right);
        return res;
    }
}

迭代

class Solution {List<Integer> res = new ArrayList<Integer>();
    Stack<TreeNode> stack = new Stack<TreeNode>();
    public List<Integer> preorderTraversal(TreeNode root) {if(root == null) return res;
        // 前序根节点间接拜访
        stack.push(root);
        while(!stack.isEmpty()) {TreeNode cur = stack.pop();
            res.add(cur.val);
            if(cur.right != null) {stack.push(cur.right);
            }

            if(cur.left != null) {stack.push(cur.left);
            }
        }
        return res;
    }
}

中序遍历

递归

class Solution {List<Integer> res = new ArrayList<Integer>();
    public List<Integer> inorderTraversal(TreeNode root) {if(root == null) return res;
        inorderTraversal(root.left);
        res.add(root.val);
        inorderTraversal(root.right);
        return res;
    }
}

迭代

class Solution {List<Integer> res = new ArrayList<Integer>();
    Stack<TreeNode> stack = new Stack<TreeNode>();
    public List<Integer> inorderTraversal(TreeNode root) {if(root == null) return res;
        // 中序 左 - 根 - 右
        while(root != null || !stack.isEmpty()) {
            // 根节点不空就间接压栈
            if(root != null) {stack.push(root);
                root = root.left;
            }else{
                // 根节点为空出栈间接拜访左节点
                root = stack.pop();
                res.add(root.val);
                root = root.right;
            }
        }
        return res;
    }
}

后序遍历

递归

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

迭代

class Solution {LinkedList<Integer> res = new LinkedList<Integer>();
    Stack<TreeNode> stack = new Stack<TreeNode>();

    public List<Integer> postorderTraversal(TreeNode root) {if(root == null) return res;
        // 根节点最初拜访先入栈
        stack.push(root);
        while(!stack.isEmpty()) {TreeNode cur = stack.pop();
            res.addFirst(cur.val);
            if(cur.left != null) {stack.push(cur.left);
            }
            if(cur.right != null) {stack.push(cur.right);
            }
            
        }

        return res;
    }
}