二叉树遍历小结

二叉树遍历小结

声明

文章均为本人技术笔记，转载请注明出处：
[1] https://segmentfault.com/u/yzwall
[2] blog.csdn.net/j_dark/

0 二叉树遍历概述

二叉树遍历：按照既定序，对每个节点仅访问一次；
二叉树非递归遍历思想：参考这篇博文，核心思想是存在重合元素的局部有序保证整体有序，由于二叉树的结构特点，二叉树中的每个节点（除根节点和叶子节点）均属于两个局部的重合元素。对于任一重合元素，保证所在两个局部遍历有序，保证实现整体遍历有序；

• 重合元素所在局部：

  • 局部全部有序，遍历该元素并出栈；

  • 局部未全部有序，将未有序局部元素全部入栈。由于栈是LIFO，局部元素按照逆序入栈；

二叉树节点TreeNode声明

public class TreeNode {
    public int val;
    public TreeNode left, right;
    public TreeNode(int val) {
        this.val = val;
        this.left = this.right = null;
    }
}

1 前序遍历

• lintcode 二叉树的前序遍历

1.1 非递归实现

public class Solution {
    private class Pair {
        public TreeNode node;
        public boolean isVisited;
        public Pair(TreeNode node, boolean isVisited) {
            this.node = node;
            this.isVisited = isVisited;
        }
    }
    
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        if (root == null) {
            return list;
        }
        
        ArrayDeque<Pair> stack = new ArrayDeque<Pair>();
        stack.push(new Pair(root, false));
        while (!stack.isEmpty()) {
            Pair top = stack.pop();
            // 重合节点完成所有局部有序，弹出
            if (top.isVisited) {
                list.add(top.node.val);
            } else {
                // reverse: right -> left -> root
                if (top.node.right != null) {
                    stack.push(new Pair(top.node.right, false));
                }               
                if (top.node.left != null) {
                    stack.push(new Pair(top.node.left, false));
                }
                stack.push(new Pair(top.node, true));
            }
        }
        return list;
    }
}

1.2 递归实现

public class Solution {
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        if (root == null) {
            return list;
        }
        traverse(list, root);
        return list;
    }
    
    private void traverse(ArrayList<Integer>list, TreeNode root) {
        if (root == null) {
            return;
        }
        list.add(root.val);
        traverse(list, root.left);
        traverse(list, root.right);
    }
}

2 中序遍历

• lintcode 二叉树的中序遍历

2.1 非递归实现

public class Solution {
    private class Pair {
        public TreeNode node;
        public boolean isVisited;
        public Pair(TreeNode node, boolean isVisited) {
            this.node = node;
            this.isVisited = isVisited;
        }
    }
    
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        if (root == null) {
            return list;
        }
        
        ArrayDeque<Pair> stack = new ArrayDeque<Pair>();
        stack.push(new Pair(root, false));
        while (!stack.isEmpty()) {
            Pair top = stack.pop();
            if (top.isVisited) {
                list.add(top.node.val);
            } else {
                // reverse: right -> root -> left
                if (top.node.right != null) {
                     stack.push(new Pair(top.node.right, false));
                }
                stack.push(new Pair(top.node, true));
                if (top.node.left != null) {
                     stack.push(new Pair(top.node.left, false));
                }
            }
        }
        return list;
    }
}

2.2 递归实现

public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        if (root == null) {
            return list;
        }
        traverse(list, root);
        return list;
    }
    
    private void traverse(ArrayList<Integer>list, TreeNode root) {
        if (root == null) {
            return;
        }
        traverse(list, root.left);
        list.add(root.val);
        traverse(list, root.right);
    }
}

3 后序遍历

• lintcode 二叉树的后序遍历

3.1 非递归实现

public class Solution {
    private class Pair {
        public TreeNode node;
        public boolean isVisited;
        public Pair(TreeNode node, boolean isVisited) {
            this.node = node;
            this.isVisited = isVisited;
        }
    }
    
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        if (root == null) {
            return list;
        }
        
        ArrayDeque<Pair> stack = new ArrayDeque<Pair>();
        stack.push(new Pair(root, false));
        while (!stack.isEmpty()) {
            Pair top = stack.pop();
            if (top.isVisited) {
                list.add(top.node.val);
            } else {
                // reverse: root -> right -> left
                stack.push(new Pair(top.node, true));
                if (top.node.right != null) {
                     stack.push(new Pair(top.node.right, false));
                }
                if (top.node.left != null) {
                     stack.push(new Pair(top.node.left, false));
                }
            }
        }

        return list;
    }
}

3.2 递归实现

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

4 层序遍历

• lintcode 二叉树的层序遍历，BFS按层遍历实现

public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayDeque<TreeNode> queue = new ArrayDeque<TreeNode>();
        ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();
        if (root == null) {
            return list;
        }
        queue.offer(root);
        while (!queue.isEmpty()) {
            int level = queue.size();
            ArrayList<Integer> levelList = new ArrayList<Integer>();
            // 按层BFS遍历
            for (int i = 0; i < level; i++) {
                TreeNode head = queue.poll();
                levelList.add(head.val);
                if (head.left != null) {
                    queue.offer(head.left);
                } 
                if (head.right != null) {
                    queue.offer(head.right);
                }
            }
            list.add(levelList);
        }
        return list;
    }
}