[LeetCode] 513. Find Bottom Left Tree Value

Given a binary tree, find the leftmost value in the last row of the tree.

Example 1:

Input:

    2
   / 
  1   3

Output:
1

Example 2:

Input:

        1
       / 
      2   3
     /   / 
    4   5   6
       /
      7

Output:
7

Note: You may assume the tree (i.e., the given root node) is not NULL.

找树左下角的值。

题意是给一个二叉树，请return最底层，最靠左边节点的值。这个题可以用BFS和DFS做。两种做法的时间和空间复杂度都是O(n)。

首先BFS比较直观，只要按层序遍历的做法一层层把node塞进queue。当遍历到最底层的时候，输出第一个塞进去的节点值即可。注意每层节点在塞入queue的时候应该是先塞右孩子再塞左孩子，原因是BFS的解法不看当前遍历到了第几层，只能通过这种方式才能保证最底层最靠左的叶子节点是最后被塞入queue的。如果按照第一个例子跑一遍，结果就是2 - 3 - 1而不是2 - 1 - 3.

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {number}
 */
var findBottomLeftValue = function (root) {
    // corner case
    if (root === null) {
        return -1;
    }

    // normal case
    let res = null;
    let queue = [root];
    while (queue.length > 0) {
        let cur = queue.shift();
        res = cur.val;
        if (cur.right) {
            queue.push(cur.right);
        }
        if (cur.left) {
            queue.push(cur.left);
        }
    }
    return res;
};

Java实现

/**
 * 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 {
    public int findBottomLeftValue(TreeNode root) {
        // corner case
        if (root == null) {
            return -1;
        }

        // normal case
        int res = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            res = cur.val;
            if (cur.right != null) {
                queue.offer(cur.right);
            }
            if (cur.left != null) {
                queue.offer(cur.left);
            }
        }
        return res;
    }
}

用DFS做的思路就稍微复杂一些。DFS的做法会用到先序遍历，会先判断左子树。很显然如果有左子树，左子树的深度会是当前深度+1；但是遍历到右子树的时候，深度不会变化。

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {number}
 */
var findBottomLeftValue = function(root) {
    let highestLevel = -1;
    let leftMost = null;
    const helper = (node, level) => {
        if (!node) return;
        if (level > highestLevel) {
            highestLevel = level;
            leftMost = node.val;
        }
        helper(node.left, level + 1);
        helper(node.right, level + 1);
    };
    helper(root, 0);
    return leftMost;
};

Java实现

/**
 * Definition for a binary tree node. public class TreeNode { int val; TreeNode
 * left; TreeNode right; TreeNode(int x) { val = x; } }
 */
class Solution {
    int res = 0;
    int height = 0;

    public int findBottomLeftValue(TreeNode root) {
        if (root == null)
            return -1;
        helper(root, 1);
        return res;
    }

    public void helper(TreeNode root, int depth) {
        if (root == null)
            return;
        if (height < depth) {
            res = root.val;
            height = dpeth;
        }
        helper(root.left, depth + 1);
        helper(root.right, depth + 1);
    }
}

LeetCode 题目总结