计算给定二叉树的所有左叶子之和.
示例：
6
/ \
4 8
/ \ / \
1 5 7 9
/
11
在这个二叉树中，有两个左叶子，分别是 1、 7和11，所以返回 19
一个节点为左叶子节点，当且仅当它是某个节点的左子节点，并且它是一个叶子结点。因此我们可以考虑对整棵树进行遍历，当我们遍历到节点时，如果它的左子节点是一个叶子结点，那么就将它的左子节点的值累加计入答案。
package com.loo;
import java.util.Stack;
public class StackLeafSum {
static class TreeNode {
TreeNode left;
TreeNode right;
int value;
TreeNode(int v) {
value = v;
}
}
public static void main(String[] args) {
// 构造示例的二叉树
TreeNode root = new TreeNode(6);
TreeNode l1 = new TreeNode(4);
TreeNode l2 = new TreeNode(1);
TreeNode l3 = new TreeNode(5);
TreeNode r1 = new TreeNode(8);
TreeNode r2 = new TreeNode(7);
TreeNode r3 = new TreeNode(9);
TreeNode r4 = new TreeNode(11);
root.left = l1;
l1.left = l2;
l1.right = l3;
root.right = r1;
r1.left = r2;
r1.right = r3;
r3.left = r4;
System.out.println(stackLeafSum(root));
}
public static int stackLeafSum(TreeNode root) {
if (root == null) {
return 0;
}
int sum = 0;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.clear();
stack.push(root);
while (stack.size()>0) {
TreeNode node = stack.pop();
if (node.left != null) {
if (isLeafNode(node.left)) {
sum += node.left.value;
} else {
stack.push(node.left);
}
}
if (node.right != null) {
stack.push(node.right);
}
}
return sum;
}
// 判断是否是叶子节点
public static boolean isLeafNode(TreeNode node) {
return node.left == null && node.right == null;
}
}